Can a struct be smaller than the sum of its components?

Question

I know that the compiler may add some padding bytes in a struct. But is it possible, when the compiler sees that we never read from a variable inside a struct, that the struct will have a smaller size than the total size of the members?

struct Foo_T
{
  int a;
  intmax_t b;
};


void bar(void)
{
  struct Foo_T foo;
  foo.a=rand();
  someFunction(foo.a);
  //i never access foo.b, only foo.a
  if(sizeof(foo)< sizeof(int)+sizeof(intmax_t))
    {
      //is it possible that we can end here?
    }
}

Answer 1

No, this is prohibited by the C standard. In C11, section 6.7.2.1 contains this statement:

15 Within a structure object, the non-bit-field members and the units in which bit-fields reside have addresses that increase in the order in which they are declared. [... ] There may be unnamed padding within a structure object, but not at its beginning.

Removing members of a struct would violate the requirement that the members have addresses that increase in the order in which they are declared.

Answer 2

No it isn't possible. When you take sizeof(foo) you expect to get at least sizeof(int) + sizeof(intmax_t). If the compiler would have given you a lesser size, it would have incorrectly affected the behavior of the program, which isn't allowed.

Suppose that you put the last member there as a place-holder "dummy", to guarantee that a reserved hardware register isn't used, or to ensure correct alignment. If the compiler would remove such a member, it would have broken the program.