Question

Are pointers on a 64-bit system still 4 byte aligned (similar to a double on a 32 bit system)? Or are they note 8 byte aligned?

For example, on a 64-bit system how big is the following data structure:

struct a {
    void* ptr;
    char myChar;
}

Would the pointer by 8 byte aligned, causing 7 bytes of padding for the character (total = 8 + 8 = 16)? Or would the pointer be 4 byte aligned (4 bytes + 4 bytes) causing 3 bytes of padding (total = 4 + 4 + 4 = 12)?

Thanks,
Ryan

Answer 1

Data alignment and packing are implementation specific, and can be usually changed from compiler settings (or even with pragmas).

However assuming you’re using default settings, on most (if not all) compilers the structure should end up being 16 bytes total. The reason is because computers reads a data chunk with size of its native word size (which is 8 bytes in 64-bit system). If it were to pad it to 4 byte offsets, the next structure would not be properly padded to 64-bit boundary. For example in case of a arr[2], the second element of the array would start at 12-byte offset, which isn’t at the native byte boundary of the machine.