Thread: C Structure Padding

Anyone know where is the reference of C Structure Padding?

I mean, how do they pad data inside a structure?

Thanks in advance

Look in the documentation for your compiler.
It's nothing to do with the standard.

The standard basically says two things
- the first element is aligned with the beginning of the structure
- the struct members are in the order written in the source code

Look in the documentation for your compiler.
It's nothing to do with the standard.

Does GCC/MSVC have it? I don't think so

By "pad" I presume you mean you want (for example) to make a 16 byte struct out of a 14 byte one. It would seem the best way to do this is with an extra char array, as it may be the only way to insure the exact number of bytes. I noticed with my compiler that if you use a short as a two byte pad, it may get further padded up to a full word. int unused:16 gives two bytes, but in this example (if you swapped char unused[2] for int unused), int unused:8 will also add 2 bytes and a just plain int unused actually adds 6 bytes (all the way to 20, according to sizeof(struct example)). I guess this is because the int is starting in the middle of a word and not on a word boundary.

Code:

struct example {
           char one[7];
           char two[7];
           char unused[2];
}

Originally Posted by MK27

By "pad" I presume you mean you want (for example) to make a 16 byte struct out of a 14 byte one. It would seem the best way to do this is with an extra char array, as it may be the only way to insure the exact number of bytes. I noticed with my compiler that if you use a short as a two byte pad, it may get further padded up to a full word. int unused:16 gives two bytes, but in this example (if you swapped char unused[2] for int unused), int unused:8 will also add 2 bytes and a just plain int unused actually adds 6 bytes (all the way to 20, according to sizeof(struct example)). I guess this is because the int is starting in the middle of a word and not on a word boundary.

Code:

struct example {
           char one[7];
           char two[7];
           char unused[2];
}

Your belief that there's not a padding byte between one and two, and between two and unused, is probably incorrect.

Edit: Actually, looking with offsetof, using int there's no padding between one and two, but there is between two and unused. (And there's only two bytes of padding, as 7+7+4=18.)

Originally Posted by tabstop

Your belief that there's not a padding byte between one and two, and between two and unused, is probably incorrect.

Edit: Actually, looking with offsetof, using int there's no padding between one and two, but there is between two and unused. (And there's only two bytes of padding, as 7+7+4=18.)

Hmm. But sizeof(struct example) will always return the correct value, correct?

(I assume these are compiler differences because this one is 16 bytes for me, which means the compiler didn't add anything. But as mentioned, if I use int unused, I get 20 bytes which I guess is 7+7+2+4).

The only way to ensure the struct has the right size is to disable padding and that is a compiler-specific option that differs between compilers.

Yes, it will return the actual size.

7+7+2+4

There must be a formula to count that Anyone know?

It would make more sense that it aligned the odd chars on a DWORD boundary, though. At least to me.