问题描述

我听说 reinterpret_cast 是实现定义的，但我不知道这到底意味着什么.你能提供一个例子来说明它是如何出错的，它出错了，使用 C-Style cast 会更好吗?

I hear that reinterpret_cast is implementation defined, but I don't know what this really means. Can you provide an example of how it can go wrong, and it goes wrong, is it better to use C-Style cast?

推荐答案

C 风格的转换并不好.

The C-style cast isn't better.

它只是按顺序尝试各种 C++ 风格的类型转换，直到找到一个有效的类型.这意味着当它像 reinterpret_cast 一样运行时，它会遇到与 reinterpret_cast 完全相同的问题.但除此之外，它还有这些问题:

It simply tries the various C++-style casts in order, until it finds one that works. That means that when it acts like a reinterpret_cast, it has the exact same problems as a reinterpret_cast. But in addition, it has these problems:

• 它可以做很多不同的事情，并且通过阅读代码并不总是清楚将调用哪种类型的转换(它的行为可能类似于 reinterpret_cast、const_cast或 static_cast，它们做的事情非常不同)
• 因此，更改周围的代码可能会改变演员表的行为
• 阅读或搜索代码时很难找到 - reinterpret_cast 很容易找到，这很好，因为 casts 很丑陋，使用时应注意.相反，通过搜索可靠地找到 C 风格的类型转换(如 (int)42.0)要困难得多

• It can do many different things, and it's not always clear from reading the code which type of cast will be invoked (it might behave like a reinterpret_cast, a const_cast or a static_cast, and those do very different things)
• Consequently, changing the surrounding code might change the behaviour of the cast
• It's hard to find when reading or searching the code - reinterpret_cast is easy to find, which is good, because casts are ugly and should be paid attention to when used. Conversely, a C-style cast (as in (int)42.0) is much harder to find reliably by searching

要回答您问题的另一部分，是的，reinterpret_cast 是实现定义的.这意味着当您使用它从 int* 转换为 float* 时，您不能保证结果指针将指向相同的地址.那部分是实现定义的.但是，如果您将结果 float* 和 reinterpret_cast 返回到 int* 中，那么您将获得原始指针.那部分是有保证的.

To answer the other part of your question, yes, reinterpret_cast is implementation-defined. This means that when you use it to convert from, say, an int* to a float*, then you have no guarantee that the resulting pointer will point to the same address. That part is implementation-defined. But if you take the resulting float* and reinterpret_cast it back into an int*, then you will get the original pointer. That part is guaranteed.

但请记住，无论您使用 reinterpret_cast 还是 C 风格的强制转换，这都是正确的:

But again, remember that this is true whether you use reinterpret_cast or a C-style cast:

int i;
int* p0 = &i;

float* p1 = (float*)p0; // implementation-defined result
float* p2 = reinterpret_cast<float*>(p0); // implementation-defined result

int* p3 = (int*)p1; // guaranteed that p3 == p0
int* p4 = (int*)p2; // guaranteed that p4 == p0
int* p5 = reinterpret_cast<int*>(p1); // guaranteed that p5 == p0
int* p6 = reinterpret_cast<int*>(p2); // guaranteed that p6 == p0

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