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 <ankurseth...@gmail.com> 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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