填充如何在结构中工作？

Question

The output of this program is 28. I don't understand how? According to me this should be 32(4+4+4+4+12)+4(to maintain the alignment)=32. Please explain the reason for displaying the output 28??

struct test{
    char c;
    int d;
    int x;
    int y;
    long double p;
    }t1;

printf("%d",sizeof(t1));

Answer 1

Maybe your "long double" is actually the same as a double (8 bytes), and if you're on a 32bit processor the alignment is 4-byte.

4+4+4+4+8 = 24

What is sizeof(long double) ?

EDIT:

I used GCC's __builtin_offset_of() and __alignof__ to investigate. The actual answer that explains the size of the struct is:

4+4+4+4+12 = 28

sizeof(long double) is 12.

No padding is necessary because __alignof__(long double) is 4 and. Interestingly, __alignof__(double) is 8.

Answer 2

On a 64 bit system sizes are char - 1 byte(its not 4 bytes) int - 4 bytes long double - 12 bytes

so total is 1+4+4+4+12+padding = 28 bytes!!

Answer 3

According to this , long doubles in gcc 32-bit mode (using gcc -m32 or with a gcc that was built to produce 32-bit output, regardless of what your platform actually is) are only 4-byte aligned. Might be good to consult the gcc manual to verify that, though.

Answer 4

That seems right to me. A field immediately following p would be 4-byte aligned, so there's no need for padding at the end of the structure.

4+4+4+4+12=28

On my system, the output is 32, but on my system, sizeof(long double) is 16. (x86_64, LLVM3).

Answer 5

The 28 sizeof output is of course correct = 1 byte for char, 3 bytes padding, 3 x 4 int and 12 bytes for long double (96 bits size but 80 bit precision), which makes in total 28 bytes.

Remember that even tho you compile on x86-64 machine, you probably compile for x86-32 machine using mingw32 and that makes a difference.

Mingw32 uses 4 byte allignement for long double .