type_info不考虑cv限定符：是这样吗？ [英] type_info doesn't account for cv qualifiers: is this right?

问题描述

这是正确的行为还是这个代码打印1的g ++ 4.5的怪癖？

Is it the correct behaviour or is it a quirk of g++4.5 that this code prints 1?

#include <iostream>
#include <typeinfo>
using namespace std;

int main(){
    struct A{};
    cout<<(typeid(A)==typeid(const A)&&typeid(A)==typeid(const volatile A)&&typeid(A)==typeid(volatile A));
}

我认为cv限定符的不同类型被视为非常不同的类型，即使更少的cv限定类型可以隐式转换为更多cv限定类型。

I thought that types differing for cv-qualifiers were threated as very distinct types, even though less cv-qualified types could be implicitly cast to more cv-qualified types.

推荐答案

typeid 根据C ++标准忽略cv限定符（取自ISO / IEC 14882：2003的§5.2.8）：

typeid ignores cv qualifiers, as per the C++ standard (taken from §5.2.8 from ISO/IEC 14882:2003) :

左值表达式的顶级cv限定符或者是typeid的操作数的type-id总是被忽略的
。 [示例：

The top-level cv-qualifiers of the lvalue expression or the type-id that is the operand of typeid are always ignored. [Example:

class D { ... };
D d1;
const D d2;

typeid(d1) == typeid(d2);       // yields true
typeid(D) == typeid(const D);   // yields true
typeid(D) == typeid(d2);        // yields true
typeid(D) == typeid(const D&);  // yields true

$ b

—end example]

所以，你看到的结果是预期的。

So, the result you're seeing is expected.

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