在C中正确地将const void指针转换为const char指针数组 [英] Casting const void pointer to array of const char pointers properly in C

问题描述

我有一个C代码看起来像这样：

I have a piece of C code that looks like this:

const char (*foo)[2] = bar();

现在 bar（）返回（const void *）。如何正确转换这个 const 指针？代码从GCC生成此警告：

Now bar() is a function that returns a (const void *). How do I properly cast this const pointer? The code produces this warning from GCC :

"initialization discards qualifiers from pointer target type".   

以下是我的一些失败尝试：

Here are some of my unsuccessful attempts:

const char (*foo)[2] = (const char *)bar();
const char (*foo)[2] = (const void **)bar();

原来的代码工作，我只是不能摆脱警告，

The original code does work, I just can't get rid of the warnings by properly casting the return value.

编辑：这已被建议：

const char (*foo)[2] = (const char (*)[2])bar();

看起来是正确的，但GCC会发出此警告：

It appears to be correct, but GCC gives this warning :

"cast discards qualifiers from pointer target type"   

这与原来的警告几乎相同。

which is nearly identical to the original warning.

编辑2：好的，我想我已经有了。这里真正的问题是（const void *）定义 bar（）。定义（const char（*）[2]）中的 const 指的是数组的元素，指向数组的指针。这个类型定义本质上是一个数组，当由 void 指针表示 const 。真正的答案是，（const void *）失去其 cons t-ness当转换为（const char（*）[2]）。

EDIT 2 : OK, I think I've got it. The real problem here is the ( const void * ) definition of bar(). The const in the definition (const char( * )[2]) refers to the elements of the array, not the pointer to the array. This type definition is essentially an array, which when represented by a void pointer is not const. The real answer is that a ( const void * ) loses its const-ness when cast to (const char ( * )[2]).

推荐答案

其他几个人已经说明了正确的转换，但它生成虚假警告。
此警告来自可能的错误 C标准，或（取决于您的解释）GCC应特别对待的情况。我相信 const 限定符可以安全和明确地提升到数组类型。您可以使用 -Wno-cast-qual 删除该警告，但当然会消除您实际关心的情况的警告。

Several others have stated the correct cast, but it generates a spurious warning. That warning comes from a possible bug in the C standard, or (depending on your interpretation) a case that GCC should treat specially. I believe the const qualifier can be safely and unambiguously lifted to the array type. You can drop that warning with -Wno-cast-qual but of course that will eliminate warnings for cases that you actually care about.

要详细说明，类型 const char（*）[2] 表示 const char 。数组没有标记为 const ，只是数组的元素。当与类型 const void * 比较时，编译器注意到后者是指向 const 的指针，其中前者不是，从而产生警告。 C标准没有办法将数组标记为 const ，即使 const 数组等价于数组 const 。

To elaborate, the type const char (*)[2] means "pointer to array (length 2) of const char". The array is not marked const, just the elements of the array. When compared to the type const void *, the compiler notices that the latter is a pointer to const, where as the former is not, thus generating the warning. The C standard provides no way to mark an array as const, even though a const array would be equivalent to an array of const.

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