4,5
statements inside sizeof() does not get executed...it will tell just size...
int a=3;
printf("%d",sizeof(a++));
here a++ will not be executed ...it will just tell size ...
Also, lets suppose it execute statements ...then sizeof(int),sizeof(node *)
...will always produce error because int and node* are not executable
statements....
On Sat, Sep 17, 2011 at 2:48 PM, prasanth n <[email protected]> wrote:
> 4,7..if we assume the sizeof(int) as 4 bytes..
>
>
> On Sat, Sep 17, 2011 at 2:32 PM, abhinav gupta <[email protected]>wrote:
>
>> Ans will be:4,5....
>>
>>
>> On Sat, Sep 17, 2011 at 2:23 PM, Sanjay Rajpal <[email protected]> wrote:
>>
>>> #include<stdio.h>
>>> int main()
>>> {
>>> int a=5;
>>> printf("Size : %d\n",sizeof( a =15/2));
>>> printf("A is %d.",a);
>>> }
>>>
>>> What will be the value of a now ? Plz explain.
>>> Sanju
>>> :)
>>>
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>>
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>
>
>
> --
> *prasanth*
>
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