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我直到凌晨 1 点才发现我的代码中的一个错误，而我的发现让我感到非常惊讶。实际代码非常复杂，涉及包含结构联合的结构联合等，但我已将问题提炼为以下简化的失败案例。

正在发生的事情是编译器 [gcc 5.4.0] 正在更改指定初始化程序的执行顺序，以匹配它们在结构中出现的顺序。只要您使用没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。查看以下代码。它表明编译器清楚地重新排序了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，但我使用每个可能的优化级别尝试了相同的代码，并且重新排序仍然发生。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。

有没有其他人见过或注意到这个问题？

这是预期/指定的行为吗？

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1 回答 1

10

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任何副作用发生的顺序。

脚注提供了进一步的说明：

特别是，评估顺序不必与子对象初始化的顺序相同。

于 2018-01-16T20:02:01.673 回答