Your code is different from the original question. You don't declare
an instance of D, C, or B. Just declaring the type does not allocate
any storage.
Don
On Dec 10, 12:29 pm, ashish mann <ashishman...@gmail.com> wrote:
> union A{
> long int y[5];
> union B{
> double g;
> union C{
> int k;
> union D{
> char ch;
> int x[5];
> };
> };
> };
> };
>
> just removed the instances of unions from the given declaration and
> tested on dev
> A 20
> B 8
> C 4
> D 20
>
> On Sat, Dec 8, 2012 at 6:39 PM, shiv narayan <narayan.shiv...@gmail.com>wrote:
>
>
>
>
>
>
>
>
>
> > but when i compile it on Dev C it gives 24..whats the reason ?
>
> > On Fri, Dec 7, 2012 at 7:19 AM, Don <dondod...@gmail.com> wrote:
>
> >> The actual size is system dependent because the language doesn't
> >> specify the size of int or long int.
> >> I'll assuming the common convention that sizeof(int)=4 and sizeof(long
> >> int)=8.
> >> The size of a union is the size of the largest element in the union.
> >> So sizeof(D) = 5*sizeof(int)=20
> >> C and B will be the same, because there is nothing bigger in any of
> >> them.
> >> sizeof(A) will be 40 which is the size of an array of eight long ints.
> >> Don
>
> >> On Dec 7, 5:42 am, zerobyzero <narayan.shiv...@gmail.com> wrote:
> >> > what will be the size of union A ,B,C and D. also please explain the
> >> logic.
>
> >> > * union A{*
> >> > * long int y[5];*
> >> > * union B{*
> >> > * double g;*
> >> > * union C{*
> >> > * int k;*
> >> > * union D{*
> >> > * char ch;*
> >> > * int x[5];*
> >> > * }s;*
> >> > * }a;*
> >> > * }b;*
> >> > * }*p;*
>
> >> --
>
> > --
> > Shiv Narayan Sharma
> > Jt. Secretary CSI-DTU
> > +919971228389
> >www.jugadengg.com
>
> > --
>
> --
> Regards,
> Aashish Mann
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