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Comment by amboar | Hacker News Reader

Use a packed struct (e.g. __attribute__((packed))). You may take a performance hit due to lack of alignment, but that's the judgement call

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__attribute__((packed)) is not part of the standard, and therefore doesn't quite answer the question. I believe the standards-compliant way of doing this is that "you can't".

Would this work

  struct S {char data, char pad0, char pad1, char pad2};
  int main() {S dataStruct; printf("%s", dataStruct.data); return 0;}

and give me a 4-byte struct or would the compiler optimize it all away and leave me with a 1-byte struct?

Since you don't actually access the size of the structure, I see no reason why it matters. Also, FWIW, your code has undefined behavior because you call printf with the wrong type.

Yes on the UB, I'm passing a "char" into a "char *", sorry about that. I don't pass lone char data that often.

Well, I chose this example specifically to ask whether the resulting memory structure, completely independent of calling a sizeof(), would still be as if I were to call a sizeof() (which then I believe would be 4 bytes large) if not accessing the total size of the struct or any of the padding members.

It makes a difference in the memory footprint, which can become important, if you're programming a device with just 64-bytes total RAM.

> performance hit due to lack of alignment

This isn't true of modern processors.

Sure you get free unaligned access for scalars on x86, but unaligned arrays are still trouble if you use SSE (which basically everyone does).

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