How can I pack a template class? - c++

Question

I have use a packed template class. According to this I should wrap it with #pragma pack as in this example:
#include <iostream>
#pragma pack(1)
template<typename X>
struct S {
int a;
X b;
};
#pragma pack()
int main()
{
std::cout << sizeof(S<char>) << "\n";
}
Using gcc-4.4.7 on my centOS6, the program output 8, which means pack(1) didn't work. Then I tried to change main() to:
int main()
{
std::cout << sizeof(S<char>) << "\n";
#pragma pack(1)
std::cout << sizeof(S<char>) << "\n";
#pragma pack()
}
Surprisingly it still didn't work. If I comment out the first line in main() then it works. So my question is how to use pack on a template class instantiation? Is there anyway to use both packed and unpacked template class in the same program?
Update: If I use gcc-5.2.1, then it output 5 correctly. Also for the following program, I get opposite results (5 with gcc-4.4.7 and 8 with 5.2.1):
template<typename X>
struct S {
int a;
X b;
};
int main()
{
#pragma pack(1)
std::cout << sizeof(S<char>) << "\n";
#pragma pack()
}
Why do gcc-4.4.7 and 5.2.1 behave differently? Which one is correct? Is there anyway to get consistent result with different versions of gcc?

Answer 1

It works for me
#include <iostream>
#pragma pack(1)
namespace Packed
{
template<typename X>
struct S {
int a;
X b;
};
template struct S<char>;
}
#pragma pack()
namespace NotPacked
{
template<typename X>
struct S {
int a;
X b;
};
template struct S<char>;
}
int main()
{
std::cout << sizeof(Packed::S<char>) << "\n";
std::cout << sizeof(NotPacked::S<char>) << "\n";
}
You can put definition of S to header (but without #pragma once) and avoid code duplication:
#pragma pack(1)
namespace Packed
{
#include "S.h"
}
#pragma pack()
namespace NotPacked
{
#include "S.h"
}