Run dis in saving the same in .C format
m getting 0...:)
#include<stdio.h>
#include<conio.h>
struct empty{};
int main()
{
struct empty e;
int x=sizeof(e);
printf("%d",x);
getch();
return 0;
}
On Wed, Jul 6, 2011 at 11:58 PM, shiv narayan <[email protected]>wrote:
> hey i am getting size of empty struct 1.
> check my code
>
> #include<iostream>
> #include<conio.h>
> using namespace std;
> struct empty{};
>
> int main()
> {
> empty e;
> int x=sizeof(e);
> cout<<x;
> getch();
> return 0;
> }
>
> when i run this i get "1" as output
>
>
> On Jul 6, 10:16 pm, Ashish Modi <[email protected]> wrote:
> > @piyush:
> > No,one can declare the variable of empty struct and access its address
> via
> > pointer. So, when you are accessing address via pointer means some memory
> is
> > allocated for that variable. But *sizeof()* operator returns *zero*??
> why???
> >
> >
> >
> >
> >
> >
> >
> >
> >
> > On Wed, Jul 6, 2011 at 10:38 PM, T3rminal <[email protected]> wrote:
> > > @ashish
> > > Most probably because empty struct in C have nothing associated with
> it.
> > > They are as good as nothing. But empty classes in C++ can have member
> > > functions. These functions need to be associated with object, having a
> > > unique address, for that class. And unique address is not possible with
> > > class of size 0 as already explained above.
> >
> > > --
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> > > "Algorithm Geeks" group.
> > > To view this discussion on the web visit
> > >https://groups.google.com/d/msg/algogeeks/-/XP8yGGz2YbEJ.
> >
> > > To post to this group, send email to [email protected].
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> >
> > --
> > With Regards
> > Ashish Modi
> > 9423721478
>
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>
--
Sanchit Mittal
Second Year Undergraduate
Computer Engineering
Delhi College of Engineering
ph- +919582414494
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