How to portably allocate space for a particular member of a union embedded in a struct

Question

Consider the following type in C11, where MyType1 and MyType2 are previously declared types:

typedef struct {
  int tag;
  union {
    MyType1 type1;
    MyType2 type2;
  }
} MyStruct;

I'd like to allocate enough memory using malloc to hold the tag attribute and type1. Can this be done in a portable way? I guess, sizeof(tag) + sizeof(type1) may not work due to alignment issues.

Can I calculate the offset of type1 from the beginning of the structure in a portable way?

If you need `struct { int tag; MyType1 type1; }` why don't you explicitly define it? It's unclear what you need this, for but it smells like something that will probably break strict aliasing rule. — user694733, Jan 16 '15 at 10:20
@user694733: I want to use pointers to MyStruct. Depending on the tag field, I need type1 or type2. — Marc, Jan 16 '15 at 10:57

score 3 · Accepted Answer · edited Jan 16 '15 at 11:40

3

You can use offsetof(), and since that will include both the size of tag and any padding, it's enough to then add the size of type1:

void *mys = malloc(offsetof(MyStruct, type1) + sizeof (MyType1));

edited Jan 16 '15 at 11:40

Klas Lindbäck

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answered Jan 16 '15 at 10:20

unwind

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score 1 · Answer 2 · answered Jan 16 '15 at 10:17

Can I calculate the offset of type1 from the beginning of the structure in a portable way?

You can probably use offsetof from stddef.h for this.

printf("Offset of type1 in the struct: %zu\n", offsetof(MyStruct, type1));

Side note: this works because you are using an "anonymous union". If you were to say union { ... } u; type1 wouldn't be a member of MyStruct.

How to portably allocate space for a particular member of a union embedded in a struct

2 Answers2