C++ 什么时候我们应该更喜欢使用两个链接的 static_cast 而不是 reinterpret_cast
首先,这不是 为什么我们要在 C++ 中使用 reinterpret_cast,而两个链接的 static_cast 可以完成它的工作?.
我知道我们甚至不能使用两个链接的 static_cast
来实现这一点的情况,reinterpret_cast
可以做什么.但是在任何情况下,我应该更喜欢两个链接的 static_cast
而不是简单且更具可读性的 reinterpret_cast
?
I know situations where we cannot use even two chained static_cast
to achieve that, what reinterpret_cast
does. But is there any situation where I should prefer a two chained static_cast
over a simple and more readable reinterpret_cast
?
推荐答案
reinterpret_cast
应该是一个巨大的闪烁符号,表示这看起来很疯狂,但我知道我在做什么.不要因为懒惰而使用它.
reinterpret_cast
should be a huge flashing symbol that says THIS LOOKS CRAZY BUT I KNOW WHAT I'M DOING. Don't use it just out of laziness.
reinterpret_cast
表示将这些位视为……" 链式静态转换不相同,因为它们可能会根据继承晶格修改其目标.
reinterpret_cast
means "treat these bits as ..." Chained static casts are not the same because they may modify their targets according to the inheritence lattice.
struct A {
int x;
};
struct B {
int y;
};
struct C : A, B {
int z;
};
C c;
A * a = &c;
int main () {
assert (reinterpret_cast <B *> (a) != static_cast <B *> (static_cast <C *> (a)));
}
如果您不是 100% 确定 a
指向 b
,请使用 dynamic_cast
它将搜索上述解决方案(尽管使用运行时成本).请记住,这可能会返回 NULL 或引发失败.
If you are not 100% sure that a
points to a b
, use dynamic_cast
which will search for the above solution (albeit with a runtime cost). Bear in mind that this may return NULL or throw on failure.
我在回想我实际使用过reinterpret_cast
的次数,实际上只有两个:
I'm trying to think of times when I've actually used reinterpret_cast
, there are really only two:
- 当一个函数正在压缩/加密任意缓冲区时,我想使用
const char *
来遍历它 if(*reinterpret_cast<uint32_t*>(array_of_4_bytes_A) < *reinterpret_cast<uint32_t*>(array_of_4_bytes_B)
或诸如此类.像这样的行会引起审查和评论.
- when a function is zipping/encrypting an arbitrary buffer and I want to use a
const char *
to traverse it if(*reinterpret_cast<uint32_t*>(array_of_4_bytes_A) < *reinterpret_cast<uint32_t*>(array_of_4_bytes_B)
or somesuch. Lines like this invite scrutiny and demand comments.
否则,如果您有一个 A*
实际上是一个 B*
,那么您可能需要一个联合.
Otherwise if you have a A*
which is really a B*
then you probably want a union.
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