为什么可以将非const引用参数绑定到临时对象？

问题描述

char f1();
void f2(char&);

struct A {};

A    f3();
void f4(A&);

int main()
{
    f2(f1()); // error C2664. This is as expected.
    f4(f3()); // OK! Why???
}

在C ++中，非const引用参数不能绑定到临时对象;在上面的代码中， f2（f1（））; 会按预期触发错误。

I have been taught that in C++ a non-const reference parameter cannot be bound to a temporary object; and in the code above, f2(f1()); triggers an error as expected.

同样的规则不适用于代码行 f4（f3（））; ？

However, why does the same rule not apply to the code line f4(f3());?

即使我注释行 f2（f1（））; ，那么包含 f4（f3（））; code>将编译时不会出现任何错误或警告。

PS: My compiler is VC++ 2013. Even if I comment the line f2(f1());, then the code containing f4(f3()); will be compiled without any errors or warnings.

更新：

说：

所以我认为这是一个VC ++的错误。我已提交错误报告至

So I think it is a bug of VC++. I have submitted a bug report to VC++ team

推荐答案

如果使用禁用语言扩展，编译器拒绝这两个调用：

If you compile with the /Za option to disable language extensions, the compiler rejects both calls:

> cl /Za test.cpp
Microsoft (R) C/C++ Optimizing Compiler Version 18.00.21005.1 for x86
Copyright (C) Microsoft Corporation.  All rights reserved.

test.cpp
test.cpp(11): error C2664: 'void f2(char &)' : cannot convert argument 1 from 'char' to 'char &'
test.cpp(12): error C2664: 'void f4(A &)' : cannot convert argument 1 from 'A' to 'A &'
        A non-const reference may only be bound to an lvalue

有几种（非常受限制的）编译器在启用语言扩展的情况下，将仍然允许非const常量引用绑定到右值表达式。我的理解是，这主要是为了避免破坏依赖这个扩展的几个巨大的遗留代码库。

There are several (very constrained) circumstances in which the compiler, with language extensions enabled, will still allow a non-const lvalue reference to bind to an rvalue expression. My understanding is that this is largely to avoid breaking several enormous legacy codebases that rely on this "extension."

（通常，不推荐使用/ Za原因，但主要是因为Windows SDK标头不能包含/ Za选项。）

(In general, use of /Za is not recommended for many reasons, but mostly because the Windows SDK headers cannot be #included with the /Za option.)

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