在c中将数组转换为结构，反之亦然 [英] Cast an array to a struct and vice-versa in c

问题描述

设为以下结构:

typedef struct { 
    int x;
    int y;
} st;

我可以将int数组投射"到结构体st:

I can "cast" an int array to a struct st:

st z;
int t[2];
t[0] = 0;
t[1] = 1;
z = *(st*)(t);
printf("%d,%d\n", z.x, z.y);

然后输出，除了:

0,1

但是我不能将一个结构体st转换为一个int数组:

But i cant cast a struct st to an int array:

st z = {0, 1};
int t[2];
t = *(int*)(&z);

因为初始化后无法分配数组.

since an array cant be assigned after initialization.

那么，我该如何实现呢?感谢您的帮助.

So, how may i achieve this ? Thanks for helping me.

推荐答案

z = *(st*)(t);

从语法上讲，此代码将编译(分配给 struct )，并且由于违反严格的别名要求，可能会产生预期或意外的结果.

Syntactically this code will compile (assignation to struct) and may give the expected or unexpected result because it violates strict aliasing requirement.

t = *(int*)(&z);

由于多种原因，这段代码在语法上是不正确的，因此编译本身将失败.首先，您无法分配给数组，其次，您正在将指向 int 的指针分配给 int 数组.更好的选择是使用 memcpy 或 union .虽然我建议按会员副本.

This piece of code is syntactically incorrect for multiple reasons and thus compilation itself would fail. First of all, you can't assign to arrays, secondly, you are assigning a pointer to int to an int array. A better bet would be to use memcpy or union instead. Though I would suggest member by member copy.

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