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int *p = NULL;
printf("%1d",p) ;

what will be the output of this above code?

Answers were Sorted based on User's Feedback



int *p = NULL; printf("%1d",p) ; what will be the output of this above code?..

Answer / vasanth

Ans : 0

Since the pointer is having NULL Address,when we try to
print like printf("%1d",*p) it will be giving the exception
hence windows will not accept NULL pointer.

Use cout<<p; then we can print NULL address 0x00000000

Note: the ans is based on VC++ compiler.

Is This Answer Correct ?    18 Yes 0 No

int *p = NULL; printf("%1d",p) ; what will be the output of this above code?..

Answer / sourisengupta

0

Is This Answer Correct ?    5 Yes 0 No

int *p = NULL; printf("%1d",p) ; what will be the output of this above code?..

Answer / ravi

undefined

Is This Answer Correct ?    0 Yes 13 No

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