enum的reinterpret_cast错误 [英] reinterpret_cast error for enum

问题描述

为什么我不能使用reinterpret_cast操作符进行这样的转换？

Why i can't use reinterpret_cast operator for such a cast?

enum Foo { bar, baz };

void foo(Foo)
{
}

int main()
{
   // foo(0); // error: invalid conversion from 'int' to 'Foo'
   // foo(reinterpret_cast<Foo>(0)); // error: invalid cast from type 'int' to type 'Foo'
   foo(static_cast<Foo>(0)); 
   foo((Foo)0);
}

推荐答案

我认为， reinterpret_cast 可以用于所有类型的转换，因为它强制任何类型转换为另一种类型，具有此转换的所有副作用。

I think that reinterpret_cast can be use for all types of casts, because it's force any type casts to another type with all side-effects of this conversion.

这是一个常见的误解。可以使用 reinterpret_cast 执行的转换在标准的5.2.10中明确列出。 int -to - 枚举和枚举 -to- int 转换不在列表中：

That is a common misconception. Conversions which can be performed with reinterpret_cast are listed explicitly in 5.2.10 of the standard. int-to-enum and enum-to-int conversions are not in the list:

• 指向整数类型的指针，只要整数是足够大的，以保持它


• nullptr_t 整数


• 整数型或枚举指向


• 函数指向不同类型的另一个函数指针的指针


• 指向另一个的对象指针不同类型的对象指针


• nullptr_t 到其他指针类型


• 指针到成员在 T1 到 T2 的另一个指针成员，如果 T1 和 T2 是对象或函数

• Pointer to integral type, so long as the integer is large enough to hold it
• nullptr_t to integer
• integral type or enum to pointer
• function pointer to another function pointer of different type
• object pointer to another object pointer of different type
• nullptr_t to other pointer type
• pointer-to-member of T1 to a different pointer-to-member of T2 in cases where both T1 and T2 are objects or functions

reinterpret_cast 通常用于告诉编译器：嘿，我知道你认为这个内存区是一个 T ，但我'你喜欢你把它解释为一个 U （其中 T 和 U 是不相关的类型）

reinterpret_cast is typically used to tell the compiler: Hey, I know you think this region of memory is a T, but I'd like you to interpret it as a U (where T and U are unrelated types).

还值得注意的是， reinterpret_cast 可以对这些位有影响：

It is also worth noting that reinterpret_cast can have effects on the bits:

5.2.10.3

[注意：reinterpret_cast执行的映射可能，或者可能不会从原始值产生一个代表性。 - 结束笔记]

[ Note: The mapping performed by reinterpret_cast might, or might not, produce a representation dif- ferent from the original value. — end note ]

C风格的转换始终可以工作，因为它包含 static_cast 在尝试。

The C-style cast always works, because it included static_cast in its attempts.

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