enum class Foo : int8_t
{
Bar1,Bar2,Bar3,Bar4,First = Bar1,Last = Bar4
};
for (Foo foo = Foo::First; foo <= Foo::Last; ++reinterpret_cast<int8_t &>(foo))
{
...
}
我知道转换为基类的引用是平凡的类的安全.但是,由于枚举类不是事件隐式转换为其底层类型,我不知道上述代码是否以及如何保证在所有编译器中都可以工作.任何线索？
解决方法
Foo& operator++( Foo& f )
{
using UT = std::underlying_type< Foo >::type;
f = static_cast< Foo >( static_cast< UT >( f ) + 1 );
return f;
}
并使用
for (Foo foo = Foo::First; foo != Foo::Last; ++foo)
{
...
}
为了回答是否允许reinterpret_cast的问题,它全部从5.2.10 / 1开始：
5.2.10 Reinterpret cast [expr.reinterpret.cast]
1 The result of the expression
reinterpret_cast<T>(v)is the result of converting the expressionvto typeT. IfTis an lvalue reference type or an rvalue reference to function type,the result is an lvalue; ifTis an rvalue reference to object type,the result is an xvalue; otherwise,the result is a prvalue and the lvalue-to-rvalue (4.1),array-to-pointer (4.2),and function-to-pointer (4.3) standard conversions are performed on the expressionv. Conversions that can be performed explicitly usingreinterpret_castare listed below. No other conversion can be performed explicitly usingreinterpret_cast.
(强调我的)
使用引用的重新解释是基于5.2.10 / 11的指针：
11 A glvalue expression of type
T1can be cast to the type “reference toT2” if an expression of type “pointer toT1” can be explicitly converted to the type “pointer toT2” using areinterpret_cast. The result refers to the same object as the source glvalue,but with the specified type. [ Note: That is,for lvalues,a reference castreinterpret_cast<T&>(x)has the same effect as the conversion*reinterpret_cast<T*>(&x)with the built-in&and*operators (and similarly forreinterpret_cast<T&&>(x)). — end note ] No temporary is created,no copy is made,and constructors (12.1) or conversion functions (12.3) are not called.
这从哪个转变了问题：
reinterpret_cast<int8_t&>(foo)
这是否合法？
*reinterpret_cast<int8_t*>(&foo)
下一站是5.2.10 / 7：
7 An object pointer can be explicitly converted to an object pointer of a different type. When a prvalue
vof type “pointer toT1” is converted to the type “pointer to cvT2”,the result isstatic_cast<cvT2*>(static_cast<cvvoid*>(v))if bothT1andT2are standard-layout types (3.9) and the alignment requirements ofT2are no stricter than those ofT1,or if either type isvoid. Converting a prvalue of type “pointer toT1” to the type “pointer toT2” (whereT1andT2are object types and where the alignment requirements ofT2are no stricter than those ofT1) and back to its original type yields the original pointer value. The result of any other such pointer conversion is unspecified.
给定3.9 / 9,int8_t和你的枚举类型是标准布局类型,问题现在转换为：
*static_cast<int8_t*>(static_cast<void*>(&foo))
这是你失败的地方. static_cast在5.2.9中定义,没有什么可以使上述合法 – 实际上5.2.9 / 5是一个明确的提示,这是非法的.其他条款没有帮助：
> 5.2.9 / 13要求T * – > void * – > T *其中T必须相同(省略cv)
> 5.2.9 / 9和5.2.9 / 10不是关于指针,而是关于值
> 5.2.9 / 11是关于类和类层次结构
> 5.2.9 / 12是关于类成员的指针
我的结论是你的代码
reinterpret_cast<int8_t&>(foo)
不合法,其行为不是由标准定义的.
另请注意,上述5.2.9 / 9和5.2.9 / 10负责将代码合法化,我在初始答案中给出了代码,您仍然可以在顶部找到该代码.
