c++ - ADL 是调用 friend 内联函数的唯一方法吗？

Question

让我们在S的声明中定义f，作为S的友元函数:

struct S
{
    friend void f() {}
};

我找不到调用 f 的方法。

那么，这样的内联友元函数真的只能用argument-dependant lookup 调用吗？ ?

struct S
{
    friend void f() {}
    friend void g(S const&) {}
} const s;

int main()
{
    // f();     // error: 'f' was not declared in this scope
    // S::f();  // error: 'f' is not a member of 'S'
    g(s);
    // S::g(s); // error: 'g' is not a member of 'S'
}

奖励:如果我想获得一个函数指针/std::function/lambda 到 g 怎么办？

Answer 1

Is it true, then, that such an inline friend function can only be called with argument-dependant lookup?

是的。如 [namespace.memdef]/3 中所述:

If a friend declaration in a non-local class first declares a class, function, class template or function template. the friend is a member of the innermost enclosing namespace. The friend declaration does not by itself make the name visible to unqualified lookup ([basic.lookup.unqual]) or qualified lookup ([basic.lookup.qual]).

由于 f 的 only 声明是它的内联定义，因此它对合格或不合格的查找不可见。然而，ADL 对此类友元函数有一个特殊规定，[basic.lookup.argdep]/4 :

When considering an associated namespace, the lookup is the same as the lookup performed when the associated namespace is used as a qualifier ([namespace.qual]) except that:

• Any namespace-scope friend functions or friend function templates declared in associated classes are visible within their respective namespaces even if they are not visible during an ordinary lookup ([class.friend]).

---

至于你的额外问题，一个 lambda 应该可以做到:

auto exposed_g = [](S const& s){ g(s); };

它将 ADL 包装到它的主体中。尽管适用于返回类型扣除的常见警告。它将是一个值(假设您不返回 void)。