用C正确铸造常量void指针为const char指针数组 [英] Casting const void pointer to array of const char pointers properly in C

问题描述

我有件C code的，看起来像这样：

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

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

现在巴（）是一个返回（常量无效*）函数。如何正确施放此常量指针？在code从GCC会产生这样的警告：

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".   

下面是我的一些不成功的尝试：

Here are some of my unsuccessful attempts:

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

原来的code不工作，我只是无法通过适当铸造返回值摆脱警告。

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

编辑：这已建议：

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

这似乎是正确的，但GCC给出了这样的警告：

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

"cast discards qualifiers from pointer target type"   

这是几乎相同的原始警告

which is nearly identical to the original warning.

编辑2：好了，我想我已经得到了它。这里真正的问题是的（常量无效*）定义栏（）。在常量定义中的（为const char（*）[2]）指的是数组的元素，不指针的阵列。这种类型的定义实际上是一个数组，再由无效指针psented $ P $时的不的常量。真正的答案是，（常量无效*）失去转换为<$ C $时，其劣势 T岬C>（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

其他几个已经指出了正确的演员阵容，但它产生的伪警告。
这一警告来自一个可能的错误 C标准，或（取决于您的跨pretation）的GCC应该特殊对待的情况。我相信常量预选赛可以安全地，毫不含糊地提升到数组类型。您可以删除该警告与 -Wno-铸QUAL 但当然，这将消除，你真正关心的情况下警告。

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，类型为const char（*）[2] 表示指针常量数组（长度） 字符。数组没有标记常量，数组的只是元素。相比于键入常量无效* 时，编译器会认为后者是一个指向常量，其中作为前者是没有，从而产生该警告。 C标准没有提供方法来标记数组作为常量，即使常量阵列将相当于一个数组为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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