Why is the “alignment” the same on 32-bit and 64-bit systems?

Question

I was wondering whether the compiler would use different padding on 32-bit and 64-bit systems, so I wrote the code below in a simple VS2019 C++ console project:

Answer 1

A struct's alignment is the size of its largest member.

That means if you have an 8-byte(64bit) member in the struct, then the struct will align to 8 bytes.

In the case that you are describing, if the compiler allows the struct to align to 4 bytes, it possibly leads to an 8-byte member lying across the cache line boundary.

Say we have a CPU that has a 16-byte cache line. Consider a struct like this:

struct Z
{
    char s;      // 1-4 byte
    __int64 i;   // 5-12 byte
    __int64 i2;  // 13-20 byte, need two cache line fetches to read this variable
};