how in the second case it is 12?????....can u tell the clear expl..

On Wed, Feb 29, 2012 at 8:39 AM, Gene <[email protected]> wrote:

> This depends on the compiler and even on the options you give the
> compiler.  The C nor C++ standards don't say.
>
> So the asker of the question hasn't give you enough information.
>
> If you assume 32-bit x86 gcc with no packing options or pragmas, I
> think shorts (which are 2 bytes long) are aligned on 2-byte
> boundaries.  Longs and ints (both 4 bytes long) are on 4-byte
> boundaries.  Chars (1 byte) can go anywhere.  If you follow these
> rules, then the first will be laid out:
>
> Field @ Offset
>
> a @ 0 // next 3 bytes are padding to reach next 4-byte boundard
> b @ 4
> c @ 8 // next 2 bytes are padding
> d @ 12
>
> so the struct will be 16 bytes in size (a long is 4 bytes).
>
> In the second case you'll have
>
> a @ 0  // next 1 byte are padding
> b @ 2
> c @ 4  // next 3 bytes are padding
> d @ 8
>
> so the struct will be 12 bytes in size.
>
> Even if you are using a 64-bit gcc (without the -m32 flag), you'll get
> an entirely different answer!
>
>
> On Feb 29, 11:13 am, Decipher <[email protected]> wrote:
> > I need some help in understanding how padding works ??
> > Please answer the following questions with proper explanations..
> >
> > struct mystruct1
> > {
> >   char a;
> >   int b;
> >   short c;
> >   long d;
> >
> > };
> >
> > struct mystruct2
> > {
> >   char a;
> >   short b;
> >   char c;
> >   long d;
> >
> > };
> >
> > What's the sizeof above 2 structures and why ??
>
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