隐式转换：const引用vs非const引用与非引用 [英] Implicit conversion : const reference vs non-const reference vs non-reference

问题描述

请考虑此代码，

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);
}

此编译精细，运行良好。但是如果我把 f（B）改为 f（B&），它不能编译。如果我写 f（const B&），它再次编译良好，运行良好。为什么是原因和理由？

This compiles fine, runs fine. But if I change f(B) to f(B&), it doesn't compile. If I write f(const B&), it again compiles fine, runs fine. Why is the reason and rationale?

摘要：

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

我想从语言规范中听到原因，理由和参考，对于这些情况。当然，函数签名本身并不正确。而 A 隐式转换为 B 和 const B& 但不会进入 B& ，这会导致编译错误。

I would like to hear reasons, rationale and reference(s) from the language specification, for each of these cases. Of course, the function signatures themselves are not incorrect. Rather A implicitly converts into B and const B&, but not into B&, and that causes the compilation error.

推荐答案

我想听取语言规范中的原因，理由和参考

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

em> C ++的设计和进化足够了？

Is The Design and Evolution of C++ sufficient?

我犯了一个严重的错误，引用由非注释[由我评论：该措辞不精确！]初始化。例如：

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

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

由于类型的不同， int& 不能引用 double 临时生成以保存由 ss 的值初始化的 int 。因此， incr（）修改了临时变量，结果未反映回调用函数 [ emphasis mine ]。

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].

想想一下：调用引用的整个点是客户端传递由函数，函数返回后，客户端必须能够观察到更改。

Think about it: The whole point of call-by-reference is that the client passes things that are changed by the function, and after the function returns, the client must be able to observe the changes.

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