1)
main()
{
char p[ ]="%d\n";
p[1] = 'c';
printf(p,65);
}
Answer:
A
Explanation:
Due to the assignment p[1] = ‘c’ the string becomes, “%c\n”. Since this string becomes the format string for printf and ASCII value of 65 is ‘A’, the same gets printed.

2)
void ( * abc( int, void ( *def) () ) ) ();

Answer:
abc is a ptr to a function which takes 2 parameters .(a). an integer variable.(b). a ptrto a funtion which returns void. the return type of the function is void.
Explanation:
Apply the clock-wise rule to find the result.


3)
main()
{
while (strcmp(“some”,”some\0”))
printf(“Strings are not equal\n”);
}
Answer:
No output
Explanation:
Ending the string constant with \0 explicitly makes no difference. So “some” and “some\0” are equivalent. So, strcmp returns 0 (false) hence breaking out of the while loop.

__________________

J.Vijayanand

1)
main()
{
char str1[] = {‘s’,’o’,’m’,’e’};
char str2[] = {‘s’,’o’,’m’,’e’,’\0’};
while (strcmp(str1,str2))
printf(“Strings are not equal\n”);
}
Answer:
“Strings are not equal”
“Strings are not equal”
….
Explanation:
If a string constant is initialized explicitly with characters, ‘\0’ is not appended automatically to the string. Since str1 doesn’t have null termination, it treats whatever the values that are in the following positions as part of the string until it randomly reaches a ‘\0’. So str1 and str2 are not the same, hence the result.

2)
main()
{
int i = 3;
for ( ;i++=0; ) printf(“%d”,i);
}

Answer:
Compiler Error: Lvalue required.
Explanation:
As we know that increment operators return rvalues and hence it cannot appear on the left hand side of an assignment operation.

3)
void main()
{
int *mptr, *cptr;
mptr = (int*)malloc(sizeof(int));
printf(“%d”,*mptr);
int *cptr = (int*)calloc(sizeof(int),1);
printf(“%d”,*cptr);
}
Answer:
garbage-value 0
Explanation:
The memory space allocated by malloc is uninitialized, whereas calloc returns the allocated memory space initialized to zeros.

__________________

J.Vijayanand

1)
void main()
{
static int i;
while(i<=10)
(i>2)?i++:i--;
printf(“%d”, i);
}
Answer:
32767
Explanation:
Since i is static it is initialized to 0. Inside the while loop the conditional operator evaluates to false, executing i--. This continues till the integer value rotates to positive value (32767). The while condition becomes false and hence, comes out of the while loop, printing the i value.

2)
main()
{
int i=10,j=20;
j = i, j?(i,j)?i:j:j;
printf("%d %d",i,j);
}

Answer:
10 10
Explanation:
The Ternary operator ( ? : ) is equivalent for if-then-else statement. So the question can be written as:
if(i,j)
{
if(i,j)
j = i;
else
j = j;
}
else
j = j;


3)
1. const char *a;
2. char* const a;
3. char const *a;
-Differentiate the above declarations.

Answer:
1. 'const' applies to char * rather than 'a' ( pointer to a constant char )
*a='F' : illegal
a="Hi" : legal

2. 'const' applies to 'a' rather than to the value of a (constant pointer to char )
*a='F' : legal
a="Hi" : illegal

3. Same as 1.

__________________

J.Vijayanand

1)
main()
{
int i=5,j=10;
i=i&=j&&10;
printf("%d %d",i,j);
}

Answer:
1 10
Explanation:
The expression can be written as i=(i&=(j&&10)); The inner expression (j&&10) evaluates to 1 because j==10. i is 5. i = 5&1 is 1. Hence the result.

2)
main()
{
int i=4,j=7;
j = j || i++ && printf("YOU CAN");
printf("%d %d", i, j);
}

Answer:
4 1
Explanation:
The boolean expression needs to be evaluated only till the truth value of the expression is not known. j is not equal to zero itself means that the expression’s truth value is 1. Because it is followed by || and true || (anything) => true where (anything) will not be evaluated. So the remaining expression is not evaluated and so the value of i remains the same.
Similarly when && operator is involved in an expression, when any of the operands become false, the whole expression’s truth value becomes false and hence the remaining expression will not be evaluated.
false && (anything) => false where (anything) will not be evaluated.

3)
main()
{
register int a=2;
printf("Address of a = %d",&a);
printf("Value of a = %d",a);
}
Answer:
Compier Error: '&' on register variable
Rule to Remember:
& (address of ) operator cannot be applied on register variables.

__________________

J.Vijayanand

What will be the output of the following code?

void main ()
{
int i = 0 , a[3] ;
a[i] = i++;
printf ("%d",a[i]) ;

}

Answer :
garbage value

Explanation:
The output for the above code would be a garbage value. In the statement a[i] = i++; the value of the variable i would get assigned first to a[i] i.e. a[0] and then the value of i would get incremented by 1. Since a[i] i.e. a[1] has not been initialized, a[i] will have a garbage value.

__________________
-Murali..

Hi

Guess what is the OP of the following statements

int i=10;

printf("%d %d %d",--i,++i,i++);


output: ????

__________________
SeeSamJagan
- Sky is not the "LIMIT", Death is not the END, There is still something beyond....

output of the above statements is

11 12 10

because printf() will execute its parameter from right to left

so i++ will print 10 & increment to 11

then ++i will increment to 12 & print 12

finally --i will decrement to 11 & print 11

hence the output is 11 12 10

__________________
SeeSamJagan
- Sky is not the "LIMIT", Death is not the END, There is still something beyond....