C / C ++联合中元素的内存位置 [英] Memory position of elements in C/C++ union

问题描述

我在C中有一个这样的工会：

I have a union in C like this:

union AUnion {
  struct CharBuf {
    char *buf;
    size_t len;
  } charbuf;
  uint8_t num;
  double fp_num;
};

我的问题是，如果给出以下内容，我可以保证吗：

My question is, can I guarantee that if given the following:

union AUnion u;

则以下条件成立：

&u == &u.num
&u == &u.fp_num
&u == &u.charbuf

即它们都始于 u

I.e they all start at the beginning of the memory segment where u is stored.

对于使用 gcc版本5.3.0 和<$编译的C程序c $ c> -std = c11 以上是正确的：

In the case of this C program compiled with gcc version 5.3.0 and -std=c11 the above is true:

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

union AUnion {
    struct CharBuf {
        char *buf;
        size_t len;
    } charbuf;
    uint8_t num;
    double fp_num;
};

int main(void)
{
    union AUnion u;
    printf("%d\n", ((void*)&u) == ((void*)&u.charbuf));
    printf("%d\n", ((void*)&u.charbuf) == ((void*)&u.num));
    printf("%d\n", ((void*)&u.num) == ((void*)&u.fp_num));
}

打印时：

1
1
1

使用相同的编译器将上述代码编译为C ++ 11，其结果与将C11编译为相同的代码。

Compiling the code above as C++11 with the same compiler results in the same output as compiling it as C11.

但这是标准化的行为吗？它是未定义的吗？我可以在大多数C编译器中依赖此行为吗？我还能在C ++编译器中期望这种行为吗？

But is this standardized behaviour? Is it undefined? Can I rely on this behaviour with most C compilers? Can I expect this behaviour with C++ compilers as well?

推荐答案

在 6.7.2.1p16 C标准保证：

工会的规模足以容纳其最大的成员。成员中最多一个的值可以随时存储在联合对象中。 指向经过适当转换的并集对象的指针指向其每个成员（或者，如果成员是位字段，则指向它所驻留的单元），反之亦然。 p>

The size of a union is sufficient to contain the largest of its members. The value of at most one of the members can be stored in a union object at any time. A pointer to a union object, suitably converted, points to each of its members (or if a member is a bit- field, then to the unit in which it resides), and vice versa.

所以，是的，您可以依赖所有成员，地址从工会开始（请注意，对于结构的第一个成员来说是相同的。）

So, yes, you can rely on all members starting at the unions address (note this is the same for the first member of a struct).

C ++标准包含类似的句子关于C样式（即仅C样式成员） union s / struct s，因为C ++允许将 union 传递给确实需要这种布局的C函数。C++标准的相关部分为9.5。

The C++ standard includes a similar sentence with respect to C-style (i.e. only C-style members) unions/structs, because C++ allows to pass unions to C functions which does require this layout.The relevant section in the C++ standard is 9.5.

但是，请注意，标准简单类型（整数，浮点数）内可能会填充位。并且它们的内部可能会有所不同（倾向）。您还可能违反严格的别名规则（C：有效类型）。

However, note there might be padding bits inside standard simple types (integers, floats). And their internal may vary (endianess). You also might violate strict aliasing rule (C: effective type).

这篇关于C / C ++联合中元素的内存位置的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！