Question

Here is a snippet:

#pragma pack(4)
struct s1
{
    char a;
    long b;
};
#pragma pack()

#pragma pack(2)
struct s2
{
    char c;
    struct s1 st1;
};
#pragma pack()


#pragma pack(2)
struct s3
{
    char a;
    long b;
};
#pragma pack()

#pragma pack(4)
struct s4
{
    char c;
    struct s3 st3;
};
#pragma pack()

I though sizeof(s4) should be 10 or 12. But it turns out to be 8.
I am using Visual C++ 6.0. Could someone tell me why?

Answer 1

#pragma pack(2)
struct s3
{
    char a;
    long b;
};
#pragma pack()

So the packing alignment of s3 is 2, and its size is 1 (alignment 1) + 1 (padding) + 4 (alignment 2) = 6.

#pragma pack(4)
struct s4
{
    char c;
    struct s3 st3;
};
#pragma pack()

The packing alignment of s4 is 4, and its size is 1 (alignment 1) + 1 (padding) + 6 (alignment 2) = 8.

Note that #pragma pack doesn’t “extra-align” anything with looser alignment requirements. It only reduces alignment, i.e. controls “packing” alignment.