工会初始化 [英] union initialisation

问题描述

我试图在全球范围内初始化工会如下面的例子：

I'm attempting to globally initialise a union as in the following example:

#include <cstdio>

typedef union {
    char t[4];
    int i;
} a;

enum {
    w = 5000,
    x,
    y,
    z
};

a temp = {w};
int main() {
    printf("%d %d %d %d %d\n", temp.t[0],temp.t[1],temp.t[2],temp.t[3],temp.i);
    return 0;
}

不过，如果您运行code，你会注意到，无论是temp.i或temp.t的[...]实际上给正确的项目我初始化工会有。我想这将是可以避免的，如果我可以手动初始化整数成员，但遗憾的是我不能。我也无法改变元素的顺序的结构（交换int和焦炭为了初始化一切正常）内 - 他们实际上是由外部库提供。
我的问题是：我怎么能设置结构的整数成员的全球的，而不是将char [4]成员（或者，在这种情况下，刚刚的char []的第一个元素）？

However, if you run the code, you'll note that neither of temp.i or temp.t[...] actually give the correct item i initialised the union with. I'd imagine this would be avoided if I could manually initialise the integer member, but unfortunately I can't. I also can't change the ordering of elements within the struct (swapping the int and char order initialises everything properly) - they're actually provided by an external library. My question is this: how can I set the integer member of the structure globally, rather than the char[4] member (or, in this case, just the first element of the char[])?

编辑：此外，有没有一个严格的-C​​ ++解决这个问题？即，一个在这里命名结构初始化不起作用（因为它没有在语言存在）？

Also, is there a strictly-c++ solution to this problem? i.e. one where named struct initialisation doesn't work (because it doesn't exist in the language)?

推荐答案

您想做到这一点：

a temp = {i: w};

这应为 GCC 和 G ++ 。

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