Why aligning of long long union member is bigger than the containing union/struct? Is this correct?

From this question one could start to believe that alignment of a union is not less than the largest alignment of it's individual members. But I have a problem with the long long type in gcc/g++. The full example can be found here, but here are the relevant parts for my question:

union ull {
  long long m;
};

struct sll {
  long long m;
};


int main() {
#define pr(v) cout << #v ": " << (v) << endl
   pr(sizeof(long long));
   pr(__alignof__(long long));
   pr(sizeof(ull));
   pr(__alignof__(ull));
   pr(sizeof(sll));
   pr(__alignof__(sll));
};

This results in the following output:

sizeof(long long): 8
__alignof__(long long): 8
sizeof(ull): 8
__alignof__(ull): 4
sizeof(sll): 8
__alignof__(sll): 4

Why is the alignment of a union member bigger than that of the containing union?

[UPDATE]

According to Keith's answer alignof is wrong here. But I test the following and it seems that alignof tells us the true. See:

union ull {
  long long m;
};
long long a;
char b;
long long c;
char d;
ull e;
int main() {
#define pr(v) cout << #v ": " << (v) << endl
   pr(size_t((void*)&b));
   pr(size_t((void*)&c));
   pr(size_t((void*)&d));
   pr(size_t((void*)&e));
   pr(size_t((void*)&c) - size_t((void*)&b));
   pr(size_t((void*)&e) - size_t((void*)&d));
};

The output:

size_t((void*)&b): 134523840
size_t((void*)&c): 134523848
size_t((void*)&d): 134523856
size_t((void*)&e): 134523860
size_t((void*)&c) - size_t((void*)&b): 8
size_t((void*)&e) - size_t((void*)&d): 4

So, the alignment of long long is 8 and alignment of union containing long long is 4 in global data. For local scope I cannot test this since it seems that compiler is free to rearrange local data - so this trick does not work. Can you comment on this?

[/UPDATE]