命名空间范围中的运算符在全局范围中隐藏了另一个 [英] Operator in namespace scope hiding another in global scope

问题描述

这是编译器错误吗?

template <typename T>
T& operator++(T& t)
{
    return t;
}

namespace asdf {

enum Foo { };
enum Bar { };

Foo& operator++(Foo& foo);

void fun()
{
    Bar bar;
    ++bar;
}

} // end namespace asdf

int main()
{
    return 0;
}

GCC 4.7错误消息是:

The GCC 4.7 error message is:

error: no match for 'operator++' in '++bar'
note: candidate is:
note: asdf::Foo& asdf::operator++(asdf::Foo&)
note: no known conversion for argument 1 from 'asdf::Bar' to 'asdf::Foo&'

如果注释掉该行，它将编译:

It compiles if you comment out the line:

Foo& operator++(Foo& foo);

推荐答案

不，这不是bug.考虑了三组并行的运算符.成员，非成员运算符和内建函数.

No that is not a bug. There are three parallel sets of operators considered. Members, non-member operators, and builtins.

通过普通的不合格+ ADL查找来查找非成员查找，而忽略所有类成员函数.因此，全局运算符被一个更近的词汇表隐藏(并且中间的成员函数不会隐藏其他非成员).

The non-member ones are looked up by normal unqualified+ADL lookup, ignoring all class member functions. Hence the global operator is hidden by a lexical more closer one (and an intervening member function wouldn't have hidden other non-members).

请注意，重载解析发生在 名称查找之后 ¹ ；在您的情况下，名称为operator++，但是没有适当的重载.

Note that overload resolution takes place after name lookup¹; in your case the name operator++ was found, but no appropriate overload.

如果Bar已被全局声明，和/或名称空间asdf中的其他运算符，则ADL(在前一种情况下)或普通的不合格查找(在后一种情况下)会将该运算符拖入.

If Bar had been declared globally, and/or the other operator in namespace asdf, ADL (in the former case) or ordinary unqualified lookup (in the latter case) would have dragged the operator in.

¹ :Overload resolution (...) takes place after name lookup has succeeded.(C ++标准)

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