c++ - 为什么可以临时调用成员函数而全局函数不能？

Question

在下面的代码中，我将 step 作为成员函数和临时值的全局函数调用。成员函数是允许的，并且可以工作，而全局函数是不允许的，因为 从类型为“kludge”的右值初始化类型为“kludge&”的非常量引用无效。

我试图从语言的角度理解为什么允许一种行为而不允许另一种行为。从技术上讲，调用和函数似乎会以相同的方式编译，或者至少可以。

#include <iostream>

struct kludge {
    int a;
    kludge() {
        a = 1;
    }

    kludge & step() {
        a++;
        std::cout << a << ",";
        return *this;
    }
};

kludge get() {
    kludge t;
    return t;
}

kludge & step( kludge & t ) {
    t.a++;
    std::cout << t.a << ",";
    return t;
}

int main() {
    get().step();
    step( get() );
}

Answer 1

您不能将右值绑定(bind)到非常量左值引用¹。这适用于 step(get()) 作为 step 的参数，它是一个非常量左值引用，不能绑定(bind)到纯右值(纯右值) get()。

但是，成员函数本身可以在每个值类别的对象参数上调用，无论是左值还是右值 - [over.match.funcs]/4 和/5:

For non-static member functions, the type of the implicit object parameter is

• “lvalue reference to cv X” for functions declared without a ref-qualifier or with the & ref-qualifier

[..]

For non-static member functions declared without a ref-qualifier, an additional rule applies:

• even if the implicit object parameter is not const-qualified, an rvalue can be bound to the parameter as long as in all other respects the argument can be converted to the type of the implicit object parameter. [ Note: The fact that such an argument is an rvalue does not affect the ranking of implicit conversion sequences (13.3.3.2). — end note ]

但是如果您使用所谓的ref-qualifiers，您可以限制对特定成员函数有效的值类别。也就是说，如果你写:

kludge & step() & { /* .. */ }

调用 get().step() 也将是错误格式的。

---

¹)
这是众所周知的事实，但这里是 [dcl.init.ref]/5，大幅缩短:

A reference to type “cv1 T1” is initialized by an expression of type “cv2 T2” as follows:

• If the reference is an lvalue reference and the initializer expression
  • is an lvalue [..]
  • has a class type (i.e., T2 is a class type), where T1 is not reference-related to T2, and can be implicitly converted to an lvalue of type “cv3 T3,” [..]
    
    
• Otherwise, the reference shall be an lvalue reference to a non-volatile const type (i.e., cv1 shall be const), or the reference shall be an rvalue reference.