if main() were a pointer, then
char c()
{
return '1';
}
int main()
{
printf("%d",sizeof(c()));
}
would also have shown 4, but it shows 1. I think there is a hidden concept.
If we change the return type of c() to void, it shows an error.
But instead of c(), we use main() with void as return type, it shows 4.
Whats happening here ?
Sanju
:)
On Tue, Sep 6, 2011 at 10:51 AM, siddharam suresh
<siddharam....@gmail.com>wrote:
> *main()* is an address/probably the function pointer,
> Thank you,
> Sid.
>
>
>
> On Tue, Sep 6, 2011 at 11:19 PM, Sanjay Rajpal <srn...@gmail.com> wrote:
>
>> #include <stdio.h>
>> #include<conio.h>
>> int main() {
>>
>> printf(" %d",sizeof(main()));
>> ... getche();
>> return 0;
>> }
>> o/p is 4..how ???
>>
>>
>> Sanju
>> :)
>>
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