C99 复杂类型二进制文件是否与具有两个成员的结构兼容？

Question

Can I use typedef struct {double r; double i;} DC; typedef struct {double r; double i;} DC; as a binary compatible data type for C99 double complex type?

My experimental C code works in my case (gcc / x86-64 / Linux). But is this universally valid ?

/* gcc -o test test.c (WORKS) */
/* gcc -O3 -march=native -o test test.c (WORKS with volatiles) */

#include <stdio.h>
#include <stdlib.h>
#include <complex.h>

typedef struct {
    double r;
    double i;
} DC;

int main()
{
    /* var */
    const int N = 3;

    volatile DC *c = malloc(N*sizeof(DC));

    volatile double complex *z;
    z = (double complex *) c;

    int i;

    /* init */
    for(i=0; i<N; i++)
    {
       c[i].r = (double) i+1;
       c[i].i = (double) (i+1) * 10;
    }
    z[1] = -0.1 -0.2*I; // THE POINT IS HERE

    /* results */
    for(i=0; i<N; i++)
    printf("i = %d, c[i].r = %f c[i].i = %f\n", i, c[i].r, c[i].i);

    /* end */
    free((void*) c);

    return 0;
}

The program's output is here:

i = 0, c[i].r = 1.000000 c[i].i = 10.000000
i = 1, c[i].r = -0.100000 c[i].i = -0.200000
i = 2, c[i].r = 3.000000 c[i].i = 30.000000

Answer 1

The pattern I have found most reliable for testing whether gcc will recognize aliasing between two types is exemplified by:

#include <complex.h>
typedef struct  { double r,i; } d2;

double test(double complex *cd, d2 *dd)
{
  if (dd->r)
    *cd = 1.0;
  return dd->r;
}

The generated code for gcc 7.1 is

.LC1:
        .long   0
        .long   1072693248
        .long   0
        .long   0
test:
        movsd   xmm0, QWORD PTR [rsi]
        ucomisd xmm0, QWORD PTR .LC0[rip]
        jp      .L4
        je      .L1
.L4:
        movsd   xmm1, QWORD PTR .LC1[rip]
        movsd   QWORD PTR [rdi], xmm1
        movsd   xmm1, QWORD PTR .LC1[rip+8]
        movsd   QWORD PTR [rdi+8], xmm1
.L1:
        rep ret
.LC0:
        .long   0
        .long   0

The return value of the function is in xmm0, which is loaded from dd->r and tested against zero. If it's non-zero, code stores a new value to *cd, but does not reload the value in xmm0.

I'm not sure for what purpose the authors of the Standard specified how complex numbers are stored given that they did not in turn make them alias-compatible with anything else, but gcc attaches more importance to the fact that the Standard doesn't mandate that it recognize aliasing than to the fact that being able to alias layout-compatible types is useful.

Answer 2

Yes, my copy of the C11 standard says (I assume C99 says the same):

Each complex type has the same representation and alignment requirements as an array type containing exactly two elements of the corresponding real type; the first element is equal to the real part, and the second element to the imaginary part, of the complex number.

Assuming no padding is added between or after members (as it most likely will not be), your struct is binary compatible with _Complex double .

Obviously the keyword is "assuming" which implies not universally valid. Hence the above text from the standard.

If you want universally valid, you'd want to use:

double val[2]; // double _Complex