深奥C指定初始化程序失败,编译器错误或功能？

Question

我只花了大约凌晨1点追踪我的代码中的错误,我发现的确让我感到惊讶.实际代码非常复杂,涉及包含结构联合等的结构的联合,但我已经将问题提炼到以下简化的失败案例中.

发生的事情是编译器[gcc 5.4.0]正在改变指定初始值设定项的执行顺序,以匹配它们在结构中出现的顺序.只要使用不具有顺序依赖性的常量或变量初始化结构,这不会导致任何问题.请查看以下代码.它表明编译器清楚地重新排序指定的初始值设定项:

#include <stdio.h>

typedef const struct {
    const size_t First_setToOne;
    const size_t Second_setToThree;
    const size_t Third_setToTwo;
    const size_t Fourth_setToFour;
} MyConstStruct;

static void Broken(void)
{
    size_t i = 0;

    const MyConstStruct myConstStruct = {
        .First_setToOne     = ++i,
        .Third_setToTwo     = ++i,
        .Second_setToThree  = ++i,
        .Fourth_setToFour   = ++i,
    };

    printf("\nBroken:\n");
    printf("First_setToOne    should be 1, is %zd\n", myConstStruct.First_setToOne   );
    printf("Second_setToThree should be 3, is %zd\n", myConstStruct.Second_setToThree);
    printf("Third_setToTwo    should be 2, is %zd\n", myConstStruct.Third_setToTwo   );
    printf("Fourth_setToFour  should be 4, is %zd\n", myConstStruct.Fourth_setToFour );
}

static void Fixed(void)
{
    size_t i = 0;

    const size_t First_setToOne     = ++i;
    const size_t Third_setToTwo     = ++i;
    const size_t Second_setToThree  = ++i;
    const size_t Fourth_setToFour   = ++i;

    const MyConstStruct myConstStruct = {
        .First_setToOne     = First_setToOne   ,
        .Third_setToTwo     = Third_setToTwo   ,
        .Second_setToThree  = Second_setToThree,
        .Fourth_setToFour   = Fourth_setToFour ,
    };

    printf("\nFixed:\n");
    printf("First_setToOne    should be 1, is %zd\n", myConstStruct.First_setToOne   );
    printf("Second_setToThree should be 3, is %zd\n", myConstStruct.Second_setToThree);
    printf("Third_setToTwo    should be 2, is %zd\n", myConstStruct.Third_setToTwo   );
    printf("Fourth_setToFour  should be 4, is %zd\n", myConstStruct.Fourth_setToFour );
}

int main (int argc, char *argv[])
{
    (void)argc;
    (void)argv;

    Broken();
    Fixed();

    return(0);
}

输出如下:

Broken:
First_setToOne    should be 1, is 1
Second_setToThree should be 3, is 2
Third_setToTwo    should be 2, is 3
Fourth_setToFour  should be 4, is 4

Fixed:
First_setToOne    should be 1, is 1
Second_setToThree should be 3, is 3
Third_setToTwo    should be 2, is 2
Fourth_setToFour  should be 4, is 4

我怀疑优化器,但我尝试使用每个可能的优化级别相同的代码,仍然会发生重新排序.所以这个问题在基础编译器中.

我有一个解决方案,所以这更像是对其他人的警告和一般性问题.

有没有人见过或注意过这个问题？

这是预期/指定的行为吗？

Answer 1

C99标准允许以任何顺序应用副作用:

6.7.8.23:未指定初始化列表表达式中出现任何副作用的顺序.

脚注进一步澄清:

特别是,评估顺序不必与子对象初始化的顺序相同.