Good collection of C n C++ interview written Questions and answers

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C Aptitude
1. void main()
int const * p=5;
printf("%d",++(*p));
Answer:
Compiler error: Cannot modify a constant value.
Explanation:
p is a pointer to a "constant integer". But we tried to change the value of the "constant integer".
2. main()
char s[ ]="man";
int i;
for(i=0;s[ i ];i++)
printf("\n%c%c%c%c",s[ i ],*(s+i),*(i+s),i[s]);
Answer:
mmmm
aaaa
nnnn
Explanation:
s[i], *(i+s), *(s+i), i[s] are all different ways of expressing the same idea. Generally array
name is the base address for that array. Here s is the base address. i is the index number/displacement from the
base address. So, indirecting it with * is same as s[i]. i[s] may be surprising. But in the case of C it is same as
3. main()
float me = 1.1;
double you = 1.1;
if(me==you)
printf("I love U");
else
printf("I hate U");
Answer:
I hate U
Explanation:
For floating point numbers (float, double, long double) the values cannot be predicted exactly.
Depending on the number of bytes, the precession with of the value represented varies. Float takes 4 bytes and
long double takes 10 bytes. So float stores 0.9 with less precision than long double.
Rule of Thumb:
Never compare or at-least be cautious when using floating point numbers with relational
operators (== , >, <, <=, >=,!= ) .
4. main()
{
static int var = 5;
printf("%d ",var--);
if(var)
main();
}
Answer:
54321
Explanation:
When static storage class is given, it is initialized once. The change in the value of a static
variable is retained even between the function calls. Main is also treated like any other ordinary function,
which can be called recursively.
5. main()
int c[ ]={2.8,3.4,4,6.7,5};
int j,*p=c,*q=c;
for(j=0;j<5;j++) {
printf(" %d ",*c);
++q;
}
for(j=0;j<5;j++){
printf(" %d ",*p);
++p;
}
Answer:
2222223465
Explanation:
Initially pointer c is assigned to both p and q. In the first loop, since only q is incremented and
not c , the value 2 will be printed 5 times. In second loop p itself is incremented. So the values 2 3 4 6 5 will be
printed.
6. main()
extern int i;
i=20;
printf("%d",i);
Answer:
Linker Error : Undefined symbol '_i'
Explanation:
extern storage class in the following declaration,
extern int i;
specifies to the compiler that the memory for i is allocated in some other program and that address will be
given to the current program at the time of linking. But linker finds that no other variable of name i is available
in any other program with memory space allocated for it. Hence a linker error has occurred .
7. main()
int i=-1,j=-1,k=0,l=2,m;
m=i++&&j++&&k++||l++;
printf("%d %d %d %d %d",i,j,k,l,m);
Answer:
00131
2 Explanation :
Logical operations always give a result of 1 or 0 . And also the logical AND (&&) operator
has higher priority over the logical OR (||) operator. So the expression ‘i++ && j++ && k++’ is executed
first. The result of this expression is 0 (-1 && -1 && 0 = 0). Now the expression is 0 || 2 which evaluates to 1
(because OR operator always gives 1 except for ‘0 || 0’ combination- for which it gives 0). So the value of m is
1. The values of other variables are also incremented by 1.
8. main()
char *p;
printf("%d %d ",sizeof(*p),sizeof(p));
Answer:
12
Explanation:
The sizeof() operator gives the number of bytes taken by its operand. P is a character pointer,
which needs one byte for storing its value (a character). Hence sizeof(*p) gives a value of 1. Since it needs two
bytes to store the address of the character pointer sizeof(p) gives 2.
9. main()
int i=3;
switch(i)
{
default:printf("zero");
case 1: printf("one");
break;
case 2:printf("two");
break;
case 3: printf("three");
break;
}
Answer :
three
Explanation :
The default case can be placed anywhere inside the loop. It is executed only when all other
cases doesn't match.
10. main()
printf("%x",-1<<4);
Answer:
fff0
Explanation :
-1 is internally represented as all 1's. When left shifted four times the least significant 4 bits are
filled with 0's.The %x format specifier specifies that the integer value be printed as a hexadecimal value.
11. main()
char string[]="Hello World";
3 display(string);
void display(char *string)
printf("%s",string);
Answer:
Compiler Error : Type mismatch in redeclaration of function display
Explanation :
In third line, when the function display is encountered, the compiler doesn't know anything
about the function display. It assumes the arguments and return types to be integers, (which is the default type).
When it sees the actual function display, the arguments and type contradicts with what it has assumed
previously. Hence a compile time error occurs.
12. main()
int c=- -2;
printf("c=%d",c);
Answer:
c=2;
Explanation:
Here unary minus (or negation) operator is used twice. Same maths rules applies, ie. minus *
minus= plus.
Note:
However you cannot give like --2. Because -- operator can only be applied to variables as a
decrement operator (eg., i--). 2 is a constant and not a variable.
13. #define int char
main()
int i=65;
printf("sizeof(i)=%d",sizeof(i));
Answer:
sizeof(i)=1
Explanation:
Since the #define replaces the string int by the macro char
14. main()
int i=10;
i=!i>14;
printf("i=%d",i);
Answer:
i=0
Explanation:
In the expression !i>14 , NOT (!) operator has more precedence than ‘ >’ symbol. ! is a unary
logical operator. !i (!10) is 0 (not of true is false). 0>14 is false (zero).
15. #include
4
main()
char s[]={'a','b','c','\n','c','\0'};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p + ++*str1-32);
Answer:
77
Explanation:
p is pointing to character '\n'. str1 is pointing to character 'a' ++*p. "p is pointing to '\n' and that is
incremented by one." the ASCII value of '\n' is 10, which is then incremented to 11. The value of ++*p is 11.
++*str1, str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98.
Now performing (11 + 98 – 32), we get 77("M");
So we get the output 77 :: "M" (Ascii is 77).
16. #include
main()
int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d----%d",*p,*q);
Answer:
SomeGarbageValue---1
Explanation:
p=&a[2][2][2] you declare only two 2D arrays, but you are trying to access the third
2D(which you are not declared) it will print garbage values. *q=***a starting address of a is assigned integer
pointer. Now q is pointing to starting address of a. If you print *q, it will print first element of 3D array.
17. #include
main()
struct xx
{
int x=3;
char name[]="hello";
};
struct xx *s;
printf("%d",s->x);
printf("%s",s->name);
Answer:
Compiler Error
Explanation:
You should not initialize variables in declaration
18. #include
main()
5
struct xx
{
int x;
struct yy
{
char s;
struct xx *p;
};
struct yy *q;
};
Answer:
Compiler Error
Explanation:
The structure yy is nested within structure xx. Hence, the elements are of yy are to be accessed
through the instance of structure xx, which needs an instance of yy to be known. If the instance is created after
defining the structure the compiler will not know about the instance relative to xx. Hence for nested structure
yy you have to declare member.
19. main()
printf("\nab");
printf("\bsi");
printf("\rha");
Answer:
hai
Explanation:
\n - newline
\b - backspace
\r - linefeed
20. main()
int i=5;
printf("%d%d%d%d%d%d",i++,i--,++i,--i,i);
Answer:
45545
Explanation:
The arguments in a function call are pushed into the stack from left to right. The evaluation is
by popping out from the stack. and the evaluation is from right to left, hence the result.
21. #define square(x) x*x
main()
int i;
i = 64/square(4);
printf("%d",i);
Answer:
64
6
Explanation:
the macro call square(4) will substituted by 4*4 so the expression becomes i = 64/4*4 . Since /
and * has equal priority the expression will be evaluated as (64/4)*4 i.e. 16*4 = 64
22. main()
char *p="hai friends",*p1;
p1=p;
while(*p!='\0') ++*p++;
printf("%s %s",p,p1);
Answer:
ibj!gsjfoet
Explanation:
++*p++ will be parse in the given order
 *p that is value at the location currently pointed by p will be taken
 ++*p the retrieved value will be incremented
 when ; is encountered the location will be incremented that is p++ will be executed
Hence, in the while loop initial value pointed by p is ‘h’, which is changed to ‘i’ by executing ++*p and pointer
moves to point, ‘a’ which is similarly changed to ‘b’ and so on. Similarly blank space is converted to ‘!’. Thus,
we obtain value in p becomes “ibj!gsjfoet” and since p reaches ‘\0’ and p1 points to p thus p1doesnot print
anything.
23. #include
#define a 10
main()
#define a 50
printf("%d",a);
Answer:
50
Explanation:
The preprocessor directives can be redefined anywhere in the program. So the most recently
assigned value will be taken.
24. #define clrscr() 100
main()
clrscr();
printf("%d\n",clrscr());
Answer:
100
Explanation:
Preprocessor executes as a seperate pass before the execution of the compiler. So textual
replacement of clrscr() to 100 occurs.The input program to compiler looks like this :
main()
{
100;
printf("%d\n",100);
}
7 Note:
100; is an executable statement but with no action. So it doesn't give any problem
25. main()
printf("%p",main);
Answer:
Some address will be printed.
Explanation:
Function names are just addresses (just like array names are addresses).
main() is also a function. So the address of function main will be printed. %p in printf specifies that the
argument is an address. They are printed as hexadecimal numbers.
main()
{
clrscr();
}
clrscr();
Answer:
No output/error
Explanation:
The first clrscr() occurs inside a function. So it becomes a function call. In the second clrscr();
is a function declaration (because it is not inside any function).
enum colors {BLACK,BLUE,GREEN}
main()
{
printf("%d..%d..%d",BLACK,BLUE,GREEN);
return(1);
}
Answer:
0..1..2
Explanation:
enum assigns numbers starting from 0, if not explicitly defined.
void main()
{
char far *farther,*farthest;
printf("%d..%d",sizeof(farther),sizeof(farthest));
}
Answer:
4..2
Explanation:
the second pointer is of char type and not a far pointer
main()
{
8 int i=400,j=300;
printf("%d..%d");
}
Answer:
400..300
Explanation:
printf takes the values of the first two assignments of the program. Any number of printf's may
be given. All of them take only the first two values. If more number of assignments given in
the program,then printf will take garbage values.
main()
{
char *p;
p="Hello";
printf("%c\n",*&*p);
}
Answer:
H
Explanation:
* is a dereference operator & is a reference operator. They can be applied any number of
times provided it is meaningful. Here p points to the first character in the string "Hello". *p
dereferences it and so its value is H. Again & references it to an address and * dereferences it
to the value H.
main()
{
int i=1;
while (i<=5)
{
printf("%d",i);
if (i>2)
goto here;
i++;
}
}
fun()
{
here:
printf("PP");
}
Answer:
Compiler error: Undefined label 'here' in function main
Explanation:
Labels have functions scope, in other words The scope of the labels is limited to functions .
The label 'here' is available in function fun() Hence it is not visible in function main.
main()
{
static char names[5][20]={"pascal","ada","cobol","fortran","perl"};
int i;
char *t;
t=names[3];
names[3]=names[4];
9 names[4]=t;
for (i=0;i<=4;i++)
printf("%s",names[i]);
}
Answer:
Compiler error: Lvalue required in function main
Explanation:
Array names are pointer constants. So it cannot be modified.
void main()
{
int i=5;
printf("%d",i++ + ++i);
}
Answer:
Output Cannot be predicted exactly.
Explanation:
Side effects are involved in the evaluation of i
void main()
{
int i=5;
printf("%d",i+++++i);
}
Answer:
Compiler Error
Explanation:
The expression i+++++i is parsed as i ++ ++ + i which is an illegal combination of operators.
#include
main()
{
int i=1,j=2;
switch(i)
{
case 1: printf("GOOD");
break;
case j: printf("BAD");
break;
}
}
Answer:
Compiler Error: Constant expression required in function main.
Explanation:
The case statement can have only constant expressions (this implies that we cannot use
variable names directly so an error).
Note:
Enumerated types can be used in case statements.
main()
{
int i;
printf("%d",scanf("%d",&i)); // value 10 is given as input here
}
10
Answer:
1
Explanation:
Scanf returns number of items successfully read and not 1/0. Here 10 is given as input which
should have been scanned successfully. So number of items read is 1.
#define f(g,g2) g##g2
main()
{
int var12=100;
printf("%d",f(var,12));
}
Answer:
100
main()
{
int i=0;
for(;i++;printf("%d",i)) ;
printf("%d",i);
}
Answer:
1
Explanation:
before entering into the for loop the checking condition is "evaluated". Here it evaluates to 0
(false) and comes out of the loop, and i is incremented (note the semicolon after the for loop).
#include
main()
{
char s[]={'a','b','c','\n','c','\0'};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p + ++*str1-32);
}
Answer:
M
Explanation:
p is pointing to character '\n'.str1 is pointing to character 'a' ++*p meAnswer:"p is pointing to
'\n' and that is incremented by one." the ASCII value of '\n' is 10. then it is incremented to 11.
the value of ++*p is 11. ++*str1 meAnswer:"str1 is pointing to 'a' that is incremented by 1 and
it becomes 'b'. ASCII value of 'b' is 98. both 11 and 98 is added and result is subtracted from
32.
i.e. (11+98-32)=77("M");
#include
main()
{
struct xx
11 {
int x=3;
char name[]="hello";
};
struct xx *s=malloc(sizeof(struct xx));
printf("%d",s->x);
printf("%s",s->name);
}
Answer:
Compiler Error
Explanation:
Initialization should not be done for structure members inside the structure declaration
#include
main()
{
struct xx
{
int x;
struct yy
{
char s;
struct xx *p;
};
struct yy *q;
};
}
Answer:
Compiler Error
Explanation:
in the end of nested structure yy a member have to be declared.
main()
{
extern int i;
i=20;
printf("%d",sizeof(i));
}
Answer:
Linker error: undefined symbol '_i'.
Explanation:
extern declaration specifies that the variable i is defined somewhere else. The compiler passes
the external variable to be resolved by the linker. So compiler doesn't find an error. During
linking the linker searches for the definition of i. Since it is not found the linker flags an error.
main()
{
printf("%d", out);
}
int out=100;
Answer:
Compiler error: undefined symbol out in function main.
Explanation:
12
The rule is that a variable is available for use from the point of declaration. Even though a is a
global variable, it is not available for main. Hence an error.
main()
{
extern out;
printf("%d", out);
}
int out=100;
Answer:
100
Explanation:
This is the correct way of writing the previous program.
main()
{
show();
}
void show()
{
printf("I'm the greatest");
}
Answer:
Compier error: Type mismatch in redeclaration of show.
Explanation:
When the compiler sees the function show it doesn't know anything about it. So the default
return type (ie, int) is assumed. But when compiler sees the actual definition of show mismatch
occurs since it is declared as void. Hence the error.
The solutions are as follows:
1. declare void show() in main() .
2. define show() before main().
3. declare extern void show() before the use of show().
main( )
{
int a[2][3][2] = {{{2,4},{7,8},{3,4}},{{2,2},{2,3},{3,4}}};
printf(“%u %u %u %d \n”,a,*a,**a,***a);
printf(“%u %u %u %d \n”,a+1,*a+1,**a+1,***a+1);
}
Answer:
100, 100, 100, 2
114, 104, 102, 3
Explanation:
The given array is a 3-D one. It can also be viewed as a 1-D array.
2 4
7
8
3
4
2
2
2
100 102 104 106 108 110 112 114 116 118 120 122
thus, for the first printf statement a, *a, **a give address of first element . since the
indirection ***a gives the value. Hence, the first line of the output.
for the second printf a+1 increases in the third dimension thus points to value at 114, *a+1
increments in second dimension thus points to 104, **a +1 increments the first dimension thus
13
points to 102 and ***a+1 first gets the value at first location and then increments it by 1.
Hence, the output.
main( )
{
int a[ ] = {10,20,30,40,50},j,*p;
for(j=0; j<5; j++)
{
printf(“%d” ,*a);
a++;
}
p = a;
for(j=0; j<5; j++)
{
printf(“%d ” ,*p);
p++;
}
}
Answer:
Compiler error: lvalue required.
Explanation:
Error is in line with statement a++. The operand must be an lvalue and may be of any of scalar
type for the any operator, array name only when subscripted is an lvalue. Simply array name is
a non-modifiable lvalue.
main( )
{
static int a[ ] = {0,1,2,3,4};
int *p[ ] = {a,a+1,a+2,a+3,a+4};
int **ptr = p;
ptr++;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
*ptr++;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
*++ptr;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
++*ptr;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
}
Answer:
111
222
333
344
Explanation:
Let us consider the array and the two pointers with some address
0
1
2
3
100
102
104
106
108
100
102
104
106
1000 1002 1004 1006 1008
14
ptr
1000
2000
After execution of the instruction ptr++ value in ptr becomes 1002, if scaling factor for integer
is 2 bytes. Now ptr – p is value in ptr – starting location of array p, (1002 – 1000) / (scaling
factor) = 1, *ptr – a = value at address pointed by ptr – starting value of array a, 1002 has a
value 102 so the value is (102 – 100)/(scaling factor) = 1, **ptr is the value stored in the
location pointed by the pointer of ptr = value pointed by value pointed by 1002 = value
pointed by 102 = 1. Hence the output of the firs printf is 1, 1, 1.
After execution of *ptr++ increments value of the value in ptr by scaling factor, so it
becomes1004. Hence, the outputs for the second printf are ptr – p = 2, *ptr – a = 2, **ptr = 2.
After execution of *++ptr increments value of the value in ptr by scaling factor, so it
becomes1004. Hence, the outputs for the third printf are ptr – p = 3, *ptr – a = 3, **ptr = 3.
After execution of ++*ptr value in ptr remains the same, the value pointed by the value is
incremented by the scaling factor. So the value in array p at location 1006 changes from 106
10 108,. Hence, the outputs for the fourth printf are ptr – p = 1006 – 1000 = 3, *ptr – a = 108 –
100 = 4, **ptr = 4.
main( )
{
char *q;
int j;
for (j=0; j<3; j++) scanf(“%s” ,(q+j));
for (j=0; j<3; j++) printf(“%c” ,*(q+j));
for (j=0; j<3; j++) printf(“%s” ,(q+j));
}
Explanation:
Here we have only one pointer to type char and since we take input in the same pointer thus
we keep writing over in the same location, each time shifting the pointer value by 1. Suppose
the inputs are MOUSE, TRACK and VIRTUAL. Then for the first input suppose the pointer
starts at location 100 then the input one is stored as
M
O
U
S
E
\0
When the second input is given the pointer is incremented as j value becomes 1, so the input is
filled in memory starting from 101.
M
T
R
A
C
K
\0
The third input starts filling from the location 102
M
T
V
I
R
T
This is the final value stored .
The first printf prints the values at the position q, q+1 and q+2 = M T V
The second printf prints three strings starting from locations q, q+1, q+2
i.e MTVIRTUAL, TVIRTUAL and VIRTUAL.
main( )
{
void *vp;
char ch = ‘g’, *cp = “goofy”;
int j = 20;
vp = &ch;
printf(“%c”, *(char *)vp);
vp = &j;
printf(“%d”,*(int *)vp);
vp = cp;
15
printf(“%s”,(char *)vp + 3);
}
Answer:
g20fy
Explanation:
Since a void pointer is used it can be type casted to any other type pointer. vp = &ch stores
address of char ch and the next statement prints the value stored in vp after type casting it to
the proper data type pointer. the output is ‘g’. Similarly the output from second printf is ‘20’.
The third printf statement type casts it to print the string from the 4 th value hence the output is
‘fy’.
main ( )
{
static char *s[ ] = {“black”, “white”, “yellow”, “violet”};
char **ptr[ ] = {s+3, s+2, s+1, s}, ***p;
p = ptr;
**++p;
printf(“%s”,*--*++p + 3);
}
Answer:
ck
Explanation:
In this problem we have an array of char pointers pointing to start of 4 strings. Then we have
ptr which is a pointer to a pointer of type char and a variable p which is a pointer to a pointer
to a pointer of type char. p hold the initial value of ptr, i.e. p = s+3. The next statement
increment value in p by 1 , thus now value of p = s+2. In the printf statement the expression is
evaluated *++p causes gets value s+1 then the pre decrement is executed and we get s+1 – 1 =
s . the indirection operator now gets the value from the array of s and adds 3 to the starting
address. The string is printed starting from this position. Thus, the output is ‘ck’.
main()
{
int i, n;
char *x = “girl”;
n = strlen(x);
*x = x[n];
for(i=0; i{
printf(“%s\n”,x);
x++;
}
}
Answer:
(blank space)
irl
rl
l
Explanation:
Here a string (a pointer to char) is initialized with a value “girl”. The strlen function returns
the length of the string, thus n has a value 4. The next statement assigns value at the nth
location (‘\0’) to the first location. Now the string becomes “\0irl” . Now the printf statement
prints the string after each iteration it increments it starting position. Loop starts from 0 to 4.
16
The first time x[0] = ‘\0’ hence it prints nothing and pointer value is incremented. The second
time it prints from x[1] i.e “irl” and the third time it prints “rl” and the last time it prints “l”
and the loop terminates.
int i,j;
for(i=0;i<=10;i++)
{
j+=5;
assert(i<5);
}
Answer:
Runtime error: Abnormal program termination.
assert failed (i<5), ,
Explanation:
asserts are used during debugging to make sure that certain conditions are satisfied. If assertion
fails, the program will terminate reporting the same. After debugging use,
#undef NDEBUG
and this will disable all the assertions from the source code. Assertion
is a good debugging tool to make use of.
main()
{
int i=-1;
+i;
printf("i = %d, +i = %d \n",i,+i);
}
Answer:
i = -1, +i = -1
Explanation:
Unary + is the only dummy operator in C. Where-ever it comes you can just ignore it just
because it has no effect in the expressions (hence the name dummy operator).
What are the files which are automatically opened when a C file is executed?
Answer:
stdin, stdout, stderr (standard input,standard output,standard error).
57) what will be the position of the file marker?
a: fseek(ptr,0,SEEK_SET);
b: fseek(ptr,0,SEEK_CUR);
Answer :
a: The SEEK_SET sets the file position marker to the starting of the file.
b: The SEEK_CUR sets the file position marker to the current position
of the file.
main()
{
char name[10],s[12];
scanf(" \"%[^\"]\"",s);
}
How scanf will execute?
Answer:
First it checks for the leading white space and discards it.Then it matches with a quotation
mark and then it reads all character upto another quotation mark.
17
What is the problem with the following code segment?
while ((fgets(receiving array,50,file_ptr)) != EOF)
;
Answer & Explanation:
fgets returns a pointer. So the correct end of file check is checking for != NULL.
main()
{
main();
}
Answer:
Runtime error : Stack overflow.
Explanation:
main function calls itself again and again. Each time the function is called its return address is
stored in the call stack. Since there is no condition to terminate the function call, the call stack
overflows at runtime. So it terminates the program and results in an error.
main()
{
char *cptr,c;
void *vptr,v;
c=10; v=0;
cptr=&c; vptr=&v;
printf("%c%v",c,v);
}
Answer:
Compiler error (at line number 4): size of v is Unknown.
Explanation:
You can create a variable of type void * but not of type void, since void is an empty type. In
the second line you are creating variable vptr of type void * and v of type void hence an error.
main()
{
char *str1="abcd";
char str2[]="abcd";
printf("%d %d %d",sizeof(str1),sizeof(str2),sizeof("abcd"));
}
Answer:
255
Explanation:
In first sizeof, str1 is a character pointer so it gives you the size of the pointer variable. In
second sizeof the name str2 indicates the name of the array whose size is 5 (including the '\0'
termination character). The third sizeof is similar to the second one.
main()
{
char not;
not=!2;
printf("%d",not);
}
Answer:
0
18
Explanation:
! is a logical operator. In C the value 0 is considered to be the boolean value FALSE, and any
non-zero value is considered to be the boolean value TRUE. Here 2 is a non-zero value so
TRUE. !TRUE is FALSE (0) so it prints 0.
#define FALSE -1
#define TRUE 1
#define NULL 0
main() {
if(NULL)
puts("NULL");
else if(FALSE)
puts("TRUE");
else
puts("FALSE");
}
Answer:
TRUE
Explanation:
The input program to the compiler after processing by the preprocessor is,
main(){
if(0)
puts("NULL");
else if(-1)
puts("TRUE");
else
puts("FALSE");
}
Preprocessor doesn't replace the values given inside the double quotes. The check by if
condition is boolean value false so it goes to else. In second if -1 is boolean value true hence
"TRUE" is printed.
main()
{
int k=1;
printf("%d==1 is ""%s",k,k==1?"TRUE":"FALSE");
}
Answer:
1==1 is TRUE
Explanation:
When two strings are placed together (or separated by white-space) they are concatenated (this
is called as "stringization" operation). So the string is as if it is given as "%d==1 is %s". The
conditional operator( ?: ) evaluates to "TRUE".
main()
{
int y;
scanf("%d",&y); // input given is 2000
if( (y%4==0 && y%100 != 0) || y%100 == 0 )
printf("%d is a leap year");
else
printf("%d is not a leap year");
}
19
Answer:
2000 is a leap year
Explanation:
An ordinary program to check if leap year or not.
#define max 5
#define int arr1[max]
main()
{
typedef char arr2[max];
arr1 list={0,1,2,3,4};
arr2 name="name";
printf("%d %s",list[0],name);
}
Answer:
Compiler error (in the line arr1 list = {0,1,2,3,4})
Explanation:
arr2 is declared of type array of size 5 of characters. So it can be used to declare the variable
name of the type arr2. But it is not the case of arr1. Hence an error.
Rule of Thumb:
#defines are used for textual replacement whereas typedefs are used for declaring new types.
int i=10;
main()
{
extern int i;
{
int i=20;
{
const volatile unsigned i=30;
printf("%d",i);
}
printf("%d",i);
}
printf("%d",i);
}
Answer:
30,20,10
Explanation:
'{' introduces new block and thus new scope. In the innermost block i is declared as,
const volatile unsigned
which is a valid declaration. i is assumed of type int. So printf prints 30. In the next block, i
has value 20 and so printf prints 20. In the outermost block, i is declared as extern, so no
storage space is allocated for it. After compilation is over the linker resolves it to global
variable i (since it is the only variable visible there). So it prints i's value as 10.
main()
{
int *j;
{
int i=10;
j=&i;
20 }
printf("%d",*j);
}
Answer:
10
Explanation:
The variable i is a block level variable and the visibility is inside that block only. But the
lifetime of i is lifetime of the function so it lives upto the exit of main function. Since the i is
still allocated space, *j prints the value stored in i since j points i.
main()
{
int i=-1;
-i;
printf("i = %d, -i = %d \n",i,-i);
}
Answer:
i = -1, -i = 1
Explanation:
-i is executed and this execution doesn't affect the value of i. In printf first you just print the
value of i. After that the value of the expression -i = -(-1) is printed.
#include
main()
{
const int i=4;
float j;
j = ++i;
printf("%d %f", i,++j);
}
Answer:
Compiler error
Explanation:
i is a constant. you cannot change the value of constant
#include
main()
{
int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d..%d",*p,*q);
}
Answer:
garbagevalue..1
Explanation:
p=&a[2][2][2] you declare only two 2D arrays. but you are trying to access the third
2D(which you are not declared) it will print garbage values. *q=***a starting address of a is
assigned integer pointer. now q is pointing to starting address of a.if you print *q meAnswer:it
will print first element of 3D array.
#include
main()
21
{
register i=5;
char j[]= "hello";
printf("%s %d",j,i);
}
Answer:
hello 5
Explanation:
if you declare i as register compiler will treat it as ordinary integer and it will take integer
value. i value may be stored either in register or in memory.
main()
{
int i=5,j=6,z;
printf("%d",i+++j);
}
Answer:
11
Explanation:
the expression i+++j is treated as (i++ + j)
struct aaa{
struct aaa *prev;
int i;
struct aaa *next;
};
main()
{
struct aaa abc,def,ghi,jkl;
int x=100;
abc.i=0;abc.prev=&jkl;
abc.next=&def;
def.i=1;def.prev=&abc;def.next=&ghi;
ghi.i=2;ghi.prev=&def;
ghi.next=&jkl;
jkl.i=3;jkl.prev=&ghi;jkl.next=&abc;
x=abc.next->next->prev->next->i;
printf("%d",x);
}
Answer:
2
Explanation:
above all statements form a double circular linked list;
abc.next->next->prev->next->i
this one points to "ghi" node the value of at particular node is 2.
struct point
{
int x;
int y;
};
struct point origin,*pp;
main()
22
{
pp=&origin;
printf("origin is(%d%d)\n",(*pp).x,(*pp).y);
printf("origin is (%d%d)\n",pp->x,pp->y);
}
Answer:
origin is(0,0)
origin is(0,0)
Explanation:
pp is a pointer to structure. we can access the elements of the structure either with arrow mark
or with indirection operator.
Note:
Since structure point is globally declared x & y are initialized as zeroes
main()
{
int i=_l_abc(10);
printf("%d\n",--i);
}
int _l_abc(int i)
{
return(i++);
}
Answer:
9
Explanation:
return(i++) it will first return i and then increments. i.e. 10 will be returned.
main()
{
char *p;
int *q;
long *r;
p=q=r=0;
p++;
q++;
r++;
printf("%p...%p...%p",p,q,r);
}
Answer:
0001...0002...0004
Explanation:
++ operator when applied to pointers increments address according to their corresponding
data-types.
main()
{
char c=' ',x,convert(z);
getc(c);
if((c>='a') && (c<='z'))
x=convert(c);
23printf("%c",x);
}
convert(z)
{
return z-32;
}
Answer:
Compiler error
Explanation:
declaration of convert and format of getc() are wrong.
main(int argc, char **argv)
{
printf("enter the character");
getchar();
sum(argv[1],argv[2]);
}
sum(num1,num2)
int num1,num2;
{
return num1+num2;
}
Answer:
Compiler error.
Explanation:
argv[1] & argv[2] are strings. They are passed to the function sum without converting it to
integer values.
# include
int one_d[]={1,2,3};
main()
{
int *ptr;
ptr=one_d;
ptr+=3;
printf("%d",*ptr);
}
Answer:
garbage value
Explanation:
ptr pointer is pointing to out of the array range of one_d.
# include
aaa() {
printf("hi");
}
bbb(){
printf("hello");
}
ccc(){
printf("bye");
}
main()
{
24
int (*ptr[3])();
ptr[0]=aaa;
ptr[1]=bbb;
ptr[2]=ccc;
ptr[2]();
}
Answer:
bye
Explanation:
ptr is array of pointers to functions of return type int.ptr[0] is assigned to address of the
function aaa. Similarly ptr[1] and ptr[2] for bbb and ccc respectively. ptr[2]() is in effect of
writing ccc(), since ptr[2] points to ccc.
#include
main()
{
FILE *ptr;
char i;
ptr=fopen("zzz.c","r");
while((i=fgetch(ptr))!=EOF)
printf("%c",i);
}
Answer:
contents of zzz.c followed by an infinite loop
Explanation:
The condition is checked against EOF, it should be checked against NULL.
main()
{
int i =0;j=0;
if(i && j++)
printf("%d..%d",i++,j);
printf("%d..%d,i,j);
}
Answer:
0..0
Explanation:
The value of i is 0. Since this information is enough to determine the truth value of the boolean
expression. So the statement following the if statement is not executed. The values of i and j
remain unchanged and get printed.
main()
{
int i;
i = abc();
printf("%d",i);
}
abc()
{
_AX = 1000;
}
Answer:
1000
25
Explanation:
Normally the return value from the function is through the information from the accumulator.
Here _AH is the pseudo global variable denoting the accumulator. Hence, the value of the
accumulator is set 1000 so the function returns value 1000.
int i;
main(){
int t;
for ( t=4;scanf("%d",&i)-t;printf("%d\n",i))
printf("%d--",t--);
}
// If the inputs are 0,1,2,3 find the o/p
Answer:
4--0
3--1
2--2
Explanation:
Let us assume some x= scanf("%d",&i)-t the values during execution
will be,
t
i
x
4
0
-4
3
1
-2
2
2
0
main(){
int a= 0;int b = 20;char x =1;char y =10;
if(a,b,x,y)
printf("hello");
}
Answer:
hello
Explanation:
The comma operator has associativity from left to right. Only the rightmost value is returned
and the other values are evaluated and ignored. Thus the value of last variable y is returned to
check in if. Since it is a non zero value if becomes true so, "hello" will be printed.
main(){
unsigned int i;
for(i=1;i>-2;i--)
printf("c aptitude");
}
Explanation:
i is an unsigned integer. It is compared with a signed value. Since the both types doesn't match,
signed is promoted to unsigned value. The unsigned equivalent of -2 is a huge value so
condition becomes false and control comes out of the loop.
In the following pgm add a stmt in the function fun such that the address of
'a' gets stored in 'j'.
main(){
int * j;
void fun(int **);
fun(&j);
}
26
void fun(int **k) {
int a =0;
/* add a stmt here*/
}
Answer:
*k = &a
Explanation:
The argument of the function is a pointer to a pointer.
What are the following notations of defining functions known as?
i.
int abc(int a,float b)
{
/* some code */
}
ii. int abc(a,b)
int a; float b;
{
/* some code*/
}
Answer:
i. ANSI C notation
ii. Kernighan & Ritche notation
main()
{
char *p;
p="%d\n";
p++;
p++;
printf(p-2,300);
}
Answer:
300
Explanation:
The pointer points to % since it is incremented twice and again decremented by 2, it points to
'%d\n' and 300 is printed.
main(){
char a[100];
a[0]='a';a[1]]='b';a[2]='c';a[4]='d';
abc(a);
}
abc(char a[]){
a++;
printf("%c",*a);
a++;
printf("%c",*a);
}
Explanation:
The base address is modified only in function and as a result a points to 'b' then after
incrementing to 'c' so bc will be printed.
func(a,b)
27
int a,b;
{
return( a= (a==b) );
}
main()
{
int process(),func();
printf("The value of process is %d !\n ",process(func,3,6));
}
process(pf,val1,val2)
int (*pf) ();
int val1,val2;
{
return((*pf) (val1,val2));
}
Answer:
The value if process is 0 !
Explanation:
The function 'process' has 3 parameters - 1, a pointer to another function 2 and 3, integers.
When this function is invoked from main, the following substitutions for formal parameters
take place: func for pf, 3 for val1 and 6 for val2. This function returns the result of the
operation performed by the function 'func'. The function func has two integer parameters. The
formal parameters are substituted as 3 for a and 6 for b. since 3 is not equal to 6, a==b returns
0. therefore the function returns 0 which in turn is returned by the function 'process'.
void main()
{
static int i=5;
if(--i){
main();
printf("%d ",i);
}
}
Answer:
0000
Explanation:
The variable "I" is declared as static, hence memory for I will be allocated for only once, as it
encounters the statement. The function main() will be called recursively unless I becomes equal to 0,
and since main() is recursively called, so the value of static I ie., 0 will be printed every time the
control is returned.
void main()
{
int k=ret(sizeof(float));
printf("\n here value is %d",++k);
}
int ret(int ret)
{
ret += 2.5;
return(ret);
}
Answer:
Here value is 7
28
Explanation:
The int ret(int ret), ie., the function name and the argument name can be the same.
Firstly, the function ret() is called in which the sizeof(float) ie., 4 is passed, after the first
expression the value in ret will be 6, as ret is integer hence the value stored in ret will have implicit
type conversion from float to int. The ret is returned in main() it is printed after and preincrement.
void main()
{
char a[]="12345\0";
int i=strlen(a);
printf("here in 3 %d\n",++i);
}
Answer:
here in 3 6
Explanation:
The char array 'a' will hold the initialized string, whose length will be counted from 0 till the
null character. Hence the 'I' will hold the value equal to 5, after the pre-increment in the printf
statement, the 6 will be printed.
void main()
{
unsigned giveit=-1;
int gotit;
printf("%u ",++giveit);
printf("%u \n",gotit=--giveit);
}
Answer:
0 65535
Explanation:
void main()
{
int i;
char a[]="\0";
if(printf("%s\n",a))
printf("Ok here \n");
else
printf("Forget it\n");
}
Answer:
Ok here
Explanation:
Printf will return how many characters does it print. Hence printing a null character
returns 1 which makes the if statement true, thus "Ok here" is printed.
void main()
{
void *v;
int integer=2;
int *i=&integer;
v=i;
printf("%d",(int*)*v);
}
29
Answer:
Compiler Error. We cannot apply indirection on type void*.
Explanation:
Void pointer is a generic pointer type. No pointer arithmetic can be done on it. Void
pointers are normally used for,
1. Passing generic pointers to functions and returning such pointers.
2. As a intermediate pointer type.
3. Used when the exact pointer type will be known at a later point of time.
void main()
{
int i=i++,j=j++,k=k++;
printf(“%d%d%d”,i,j,k);
}
Answer:
Garbage values.
Explanation:
An identifier is available to use in program code from the point of its declaration.
So expressions such as i = i++ are valid statements. The i, j and k are automatic variables and
so they contain some garbage value. Garbage in is garbage out (GIGO).
void main()
{
static int i=i++, j=j++, k=k++;
printf(“i = %d j = %d k = %d”, i, j, k);
}
Answer:
i=1j=1k=1
Explanation:
Since static variables are initialized to zero by default.
void main()
{
while(1){
if(printf("%d",printf("%d")))
break;
else
continue;
}
}
Answer:
Garbage values
Explanation:
The inner printf executes first to print some garbage value. The printf returns no of characters
printed and this value also cannot be predicted. Still the outer printf prints something and so
returns a non-zero value. So it encounters the break statement and comes out of the while
statement.
main()
{
unsigned int i=10;
while(i-->=0)
30
printf("%u ",i);
}
Answer:
10 9 8 7 6 5 4 3 2 1 0 65535 65534…..
Explanation:
Since i is an unsigned integer it can never become negative. So the expression i-- >=0 will
always be true, leading to an infinite loop.
#include
main()
{
int x,y=2,z,a;
if(x=y%2) z=2;
a=2;
printf("%d %d ",z,x);
}
Answer:
Garbage-value 0
Explanation:
The value of y%2 is 0. This value is assigned to x. The condition reduces to if (x) or in other
words if(0) and so z goes uninitialized.
Thumb Rule: Check all control paths to write bug free code.
main()
{
int a[10];
printf("%d",*a+1-*a+3);
}
Answer:
4
Explanation:
*a and -*a cancels out. The result is as simple as 1 + 3 = 4 !
#define prod(a,b) a*b
main()
{
int x=3,y=4;
printf("%d",prod(x+2,y-1));
}
Answer:
10
Explanation:
The macro expands and evaluates to as:
x+2*y-1 => x+(2*y)-1 => 10
main()
{
unsigned int i=65000;
while(i++!=0);
printf("%d",i);
}
Answer:
31
1
Explanation:
Note the semicolon after the while statement. When the value of i becomes 0 it comes out of
while loop. Due to post-increment on i the value of i while printing is 1.
main()
{
int i=0;
while(+(+i--)!=0)
i-=i++;
printf("%d",i);
}
Answer:
-1
Explanation:
Unary + is the only dummy operator in C. So it has no effect on the expression and now the
while loop is, while(i--!=0) which is false and so breaks out of while loop. The value –1 is
printed due to the post-decrement operator.
main()
{
float f=5,g=10;
enum{i=10,j=20,k=50};
printf("%d\n",++k);
printf("%f\n",f<<2);
printf("%lf\n",f%g);
printf("%lf\n",fmod(f,g));
}
Answer:
Line no 5: Error: Lvalue required
Line no 6: Cannot apply leftshift to float
Line no 7: Cannot apply mod to float
Explanation:
Enumeration constants cannot be modified, so you cannot apply ++.
Bit-wise operators and % operators cannot be applied on float values.
fmod() is to find the modulus values for floats as % operator is for ints.
main()
{
int i=10;
void pascal f(int,int,int);
f(i++,i++,i++);
printf(" %d",i);
}
void pascal f(integer :i,integer:j,integer :k)
{
write(i,j,k);
}
Answer:
Compiler error: unknown type integer
Compiler error: undeclared function write
Explanation:
32
Pascal keyword doesn’t mean that pascal code can be used. It means that the function follows
Pascal argument passing mechanism in calling the functions.
void pascal f(int i,int j,int k)
{
printf(“%d %d %d”,i, j, k);
}
void cdecl f(int i,int j,int k)
{
printf(“%d %d %d”,i, j, k);
}
main()
{
int i=10;
f(i++,i++,i++);
printf(" %d\n",i);
i=10;
f(i++,i++,i++);
printf(" %d",i);
}
Answer:
10 11 12 13
12 11 10 13
Explanation:
Pascal argument passing mechanism forces the arguments to be called from left to right. cdecl
is the normal C argument passing mechanism where the arguments are passed from right to left.
112). What is the output of the program given below
main()
{
signed char i=0;
for(;i>=0;i++) ;
printf("%d\n",i);
}
Answer
-128
Explanation
Notice the semicolon at the end of the for loop. THe initial value of the i is set to 0.
The inner loop executes to increment the value from 0 to 127 (the positive range of
char) and then it rotates to the negative value of -128. The condition in the for loop
fails and so comes out of the for loop. It prints the current value of i that is -128.
113) main()
{
unsigned char i=0;
for(;i>=0;i++) ;
printf("%d\n",i);
}
Answer
infinite loop
Explanation
33
The difference between the previous question and this one is that the char is declared to be
unsigned. So the i++ can never yield negative value and i>=0 never becomes false so that it can come
out of the for loop.
114) main()
{
char i=0;
for(;i>=0;i++) ;
printf("%d\n",i);
}
Answer:
Behavior is implementation dependent.
Explanation:
The detail if the char is signed/unsigned by default is implementation dependent. If the
implementation treats the char to be signed by default the program will print –128 and
terminate. On the other hand if it considers char to be unsigned by default, it goes to infinite
loop.
Rule:
You can write programs that have implementation dependent behavior. But dont write
programs that depend on such behavior.
115) Is the following statement a declaration/definition. Find what does it mean?
int (*x)[10];
Answer
Definition.
x is a pointer to array of(size 10) integers.
Apply clock-wise rule to find the meaning of this definition.
116). What is the output for the program given below
typedef enum errorType{warning, error, exception,}error;
main()
{
error g1;
g1=1;
printf("%d",g1);
}
Answer
Compiler error: Multiple declaration for error
Explanation
The name error is used in the two meanings. One means that it is a enumerator
constant with value 1. The another use is that it is a type name (due to typedef) for enum
errorType. Given a situation the compiler cannot distinguish the meaning of error to know in
what sense the error is used:
error g1;
g1=error;
// which error it refers in each case?
When the compiler can distinguish between usages then it will not issue error (in pure
technical terms, names can only be overloaded in different namespaces).
Note: the extra comma in the declaration,
34
enum errorType{warning, error, exception,}
is not an error. An extra comma is valid and is provided just for programmer’s convenience.
117)
typedef struct error{int warning, error, exception;}error;
main()
{
error g1;
g1.error =1;
printf("%d",g1.error);
}
Answer
1
Explanation
The three usages of name errors can be distinguishable by the compiler at any instance, so
valid (they are in different namespaces).
Typedef struct error{int warning, error, exception;}error;
This error can be used only by preceding the error by struct kayword as in:
struct error someError;
typedef struct error{int warning, error, exception;}error;
This can be used only after . (dot) or -> (arrow) operator preceded by the variable name as in :
g1.error =1;
printf("%d",g1.error);
typedef struct error{int warning, error, exception;}error;
This can be used to define variables without using the preceding struct keyword as in:
error g1;
Since the compiler can perfectly distinguish between these three usages, it is perfectly legal and valid.
Note
This code is given here to just explain the concept behind. In real programming don’t use such
overloading of names. It reduces the readability of the code. Possible doesn’t mean that we should use
it!
118)
#ifdef something
int some=0;
#endif
main()
{
int thing = 0;
printf("%d %d\n", some ,thing);
}
Answer:
Compiler error : undefined symbol some
Explanation:
This is a very simple example for conditional compilation. The name something is not
already known to the compiler making the declaration
int some = 0;
effectively removed from the source code.
119)
#if something == 0
35
int some=0;
#endif
main()
{
int thing = 0;
printf("%d %d\n", some ,thing);
}
Answer
00
Explanation
This code is to show that preprocessor expressions are not the same as the ordinary
expressions. If a name is not known the preprocessor treats it to be equal to zero.
120). What is the output for the following program
main()
{
int arr2D[3][3];
printf("%d\n", ((arr2D==* arr2D)&&(* arr2D == arr2D[0])) );
}
Answer
1
Explanation
This is due to the close relation between the arrays and pointers. N dimensional arrays
are made up of (N-1) dimensional arrays.
arr2D is made up of a 3 single arrays that contains 3 integers each .
arr2D
arr2D[1]
arr2D[2]
arr2D[3]
The name arr2D refers to the beginning of all the 3 arrays. *arr2D refers to the start of
the first 1D array (of 3 integers) that is the same address as arr2D. So the expression
(arr2D == *arr2D) is true (1).
Similarly, *arr2D is nothing but *(arr2D + 0), adding a zero doesn’t change the
value/meaning. Again arr2D[0] is the another way of telling *(arr2D + 0). So the
expression (*(arr2D + 0) == arr2D[0]) is true (1).
Since both parts of the expression evaluates to true the result is true(1) and the same is
printed.
121) void main()
{
if(~0 == (unsigned int)-1)
printf(“You can answer this if you know how values are represented in memory”);
}
Answer
You can answer this if you know how values are represented in memory
Explanation
36
~ (tilde operator or bit-wise negation operator) operates on 0 to produce all ones to fill
the space for an integer. –1 is represented in unsigned value as all 1’s and so both are
equal.
122) int swap(int *a,int *b)
{
*a=*a+*b;*b=*a-*b;*a=*a-*b;
}
main()
{
int x=10,y=20;
swap(&x,&y);
printf("x= %d y = %d\n",x,y);
}
Answer
x = 20 y = 10
Explanation
This is one way of swapping two values. Simple checking will help understand this.
123)
main()
{
char *p = “ayqm”;
printf(“%c”,++*(p++));
}
Answer:
b
124)
main()
{
int i=5;
printf("%d",++i++);
}
Answer:
Compiler error: Lvalue required in function main
Explanation:
++i yields an rvalue. For postfix ++ to operate an lvalue is required.
125)
main()
{
char *p = “ayqm”;
char c;
c = ++*p++;
printf(“%c”,c);
}
Answer:
b
Explanation:
There is no difference between the expression ++*(p++) and ++*p++. Parenthesis just
works as a visual clue for the reader to see which expression is first evaluated.
126)
int aaa() {printf(“Hi”);}
int bbb(){printf(“hello”);}
37
iny ccc(){printf(“bye”);}
main()
{
int ( * ptr[3]) ();
ptr[0] = aaa;
ptr[1] = bbb;
ptr[2] =ccc;
ptr[2]();
}
Answer:
bye
Explanation:
int (* ptr[3])() says that ptr is an array of pointers to functions that takes no arguments and
returns the type int. By the assignment ptr[0] = aaa; it means that the first function pointer in
the array is initialized with the address of the function aaa. Similarly, the other two array
elements also get initialized with the addresses of the functions bbb and ccc. Since ptr[2]
contains the address of the function ccc, the call to the function ptr[2]() is same as calling
ccc(). So it results in printing "bye".