指针到常量用C转换 [英] pointer-to-const conversion in C

问题描述

以下code编译没有对海湾合作委员会警告，但给出了在Visual Studio 2005中的警告。

The following code compiles without warning on GCC but gives a warning in Visual Studio 2005.

const void * x = 0;
char * const * p = x;

x指向未知类型的常量对象，p指向一个常量字符指针。为什么要分配至P导致一个警告？

x points to a constant object of unknown type, and p points to a constant pointer to char. Why should the assignment to p result in a warning?

再次，这是C，而不是C ++。谢谢你。

Again, this is C, not C++. Thanks.

推荐答案

这是因为当你做一个类型的点的指针为另一种类型，有时，这样做是无意（错误），所以编译器警告你一下吧。

It happens because when you make a pointer of one type point to another type, sometimes it is done unintentionally (bug), so the compiler warns you about it.

因此​​，为了告诉你真正的意图去做，你必须做明确的铸造编译器，像这样的：

So, in order to tell the compiler that you actually intent to do it, you have to do explicit casting, like this:

        const void * x = 0;
        char * const * p = (char * const * )x;

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P.S。在我写的第一个地方的大部分的时间的是无意中做的，但AndreyT让我重新考虑它的时候，他理所当然地说，void *的专门为此目的存在。

P.S. At the first place I wrote "most of the times is done unintentionally", but AndreyT made me reconsider it when he rightfully said that void * exists specifically for that purpose.

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