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?
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:
reinterpret_cast
、const_cast
或 static_cast
,它们做的事情非常不同)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reinterpret_cast
, a const_cast
or a static_cast
, and those do very different things)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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