sizeof an inline struct declaration

Question

If there's a structure with a pointer to a struct declared within it, how do you determine the size of the sub-struct?

typedef struct myStruct {
  int member;
  struct subStruct {
    int a;
    int b;
  } *subStruct_t;
} myStruct_t;

How do you allocate space for the subStruct_t pointer? I was thinking something along the lines of

myStruct_t M;
M.subStruct_t = calloc(1,sizeof(myStruct_t.subStruct_t);

but it obviously doesn't work. Any ideas?

Answer 1

M.subStruct_t = calloc(1,sizeof(*M.subStruct_t));

Note: allocate for the size of the structure, not for the pointer

Answer 2

In C, inner structs get placed in the global namespace, so you can simply use sizeof(struct subStruct). In C++, you have to use the scope resolution operator ::, so you would instead say sizeof(myStruct::subStruct).

You can also just use the name of the dereferenced variable -- the operands to sizeof are not evaluated -- so sizeof(*M.subStruct_t) would also work.

One piece of advice: do not name your struct member with the _t suffix. The _t suffix should be used for types, not for variables/members. Furthermore, POSIX reserves all identifiers with the _t suffix (see section 2.2.2 of the POSIX.1-2008 spec), so you should not name your own types with _t.