c++ - 带有 const 参数的友元函数

Question

我知道要创建一个友元函数，友元函数应该在封闭范围内显式声明或接受其类的参数。但是，这似乎是一个警告，我无法理解。为什么调用f1(99)不工作？

class X { 
  public:
    X(int i) {
      std::cout << "Ctor called" << std::endl;
    }
    friend int f1(X&);
    friend int f2(const X&);
    friend int f3(X);
};
int f1(X& a) {
  std::cout << "non-const called" << std::endl;
}
int f2(const X& a) {
  std::cout << "const called" << std::endl;
}
int f3(X a) {
  std::cout << "object called" << std::endl;
}   

int main() {
  f1(99);
  f2(99);
  f3(99);
}

Answer 1

您正在使用参数 99 调用函数;但所有函数都需要 X . 99可以转换为 X隐含地，但转换后的 X是一个临时对象，不能绑定(bind)到非常量的左值引用。

临时对象可以绑定(bind)到 lvalue-reference to const (以及从 C++11 开始的右值引用)，然后是 f2(99);作品。并且可以复制到f3的参数中，然后 f3(99)也可以。

The effects of reference initialization are:

• Otherwise, if the reference is lvalue reference to a non-volatile const-qualified type or rvalue reference (since C++11):

  • Otherwise, object is implicitly converted to T. The reference is bound to the result of the conversion (after materializing a temporary) (since C++17). If the object (or, if the conversion is done by user-defined conversion, the result of the conversion function) is of type T or derived from T, it must be equally or less cv-qualified than T, and, if the reference is an rvalue reference, must not be an lvalue (since C++11).