c++ - 编译器看不到在类中定义的友元函数

Question

这个问题在这里已经有了答案:





What's the scope of inline friend functions?

(6 个回答)


去年关闭。




在编译以下代码时:

class A {
    A() = default;
public:
    friend A getA() {
        return A();
    }
};

int main()
{
    A a = getA();
}

编译器给了我一个错误:

'getA' was not declared in this scope

这是为什么？

Answer 1

因为friend function getA无法通过名称查找找到。

(强调我的)

A name first declared in a friend declaration within class or class template X becomes a member of the innermost enclosing namespace of X, but is not visible for lookup (except argument-dependent lookup that considers X) unless a matching declaration at the namespace scope is provided

getA找不到 ADL或者，(它没有参数)。您需要在命名空间范围内提供声明。例如

class A;
A getA();
class A {
    A() = default;
public:
    friend A getA() {
        return A();
    }
};