class with friend 而不是 forward declaration,: 哪个编译器是正确的

Question

我有这个简单的 C++ 程序：

#include <iostream>

struct obj {
  friend int f(int);

  void m(int x) { std::cout << "f(" << x << ") = " << f(x) << std::endl; }
};

int main() {
  obj o;
  o.m(21);
}


int f(int x) {
  return 2*x;
}

如果我用 GNU C++ 编译器 g++ 编译，我得到错误 prog.cpp:7:55: error: 'f' was not declared in this范围

但是，如果我用 cl（和 /W4）编译它，它会编译并执行得很好。

我不确定哪个编译器是正确的。

Answer 1

来自 friend declaration 上的 cppreference：

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

提供了全局范围内的匹配声明，但仅在 obj::m() 的定义之后。我认为此时拒绝对 f 的调用是正确的。你可以看到效果，例如这与 gcc 和 clang,

一起编译
int f(int); class obj { /* as before... */ };

还有这个：

struct obj { friend int f(int); void m(int x); }; int f(int); void obj::m(int x) { std::cout << "f(" << x << ") = " << f(x) << std::endl; }

class with friend 而不是 forward declaration,: 哪个编译器是正确的

class with friend rather than forward declaration,: which compiler is correct

c++

gcc

cl