c++ - 隐式转换 : const reference vs non-const reference vs non-reference

Question

考虑这段代码，

struct A {};
struct B {  B(const A&) {} };
void f(B)
{
    cout << "f()"<<endl;
}
void g(A &a)
{
    cout << "g()" <<endl;
    f(a); //a is implicitly converted into B.
}
int main()
{
    A a;
    g(a);
}

这compiles fine ，运行良好。但是，如果我将 f(B) 更改为 f(B&)，它 doesn't compile .如果我写f(const B&)，它又是compiles fine ，运行良好。原因和道理是什么？

总结:

void f(B);         //okay
void f(B&);        //error
void f(const B&);  //okay

对于每种情况，我想听听语言规范中的原因、基本原理和引用资料。当然，函数签名本身并没有错。 A 隐式转换为 B 和 const B&，但不会转换为 B&，这会导致编译错误。

Answer 1

I would like to hear reasons, rationale and reference(s) from the language specification

C++ 的设计和演进是否足够？

I made one serious mistake, though, by allowing a non-const reference to be initialized by a non-lvalue [comment by me: that wording is imprecise!]. For example:

void incr(int& rr) { ++rr; }

void g()
{
    double ss = 1;
    incr(ss);    // note: double passed, int expected
                 // (fixed: error in release 2.0)
}

Because of the difference in type the int& cannot refer to the double passed so a temporary was generated to hold an int initialized by ss's value. Thus, incr() modified the temporary, and the result wasn't reflected back to the calling function [emphasis mine].

想想看:call-by-reference 的全部意义在于客户端传递被函数改变的东西，函数返回后，客户端必须能够观察到变化.