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