引用和const引用之间的差异作为函数参数？ [英] Difference between reference and const reference as function parameter?

问题描述

这里是一个简单的C ++代码片段：

Here is a simple snippet of C++ code:

A foo(){
  A a; // create a local A object
  return a;
}

void bar(const A & a_r){

}

bar(foo());

为什么函数bar的参数必须是一个const引用，而不仅仅是一个引用？

Why does the argument of function bar have to be a const reference,not just a reference?

Edit1：我知道引用是为了避免复制开销。和const是只读的。但这里我必须使它成为一个const引用，否则如果我删除const，g ++会给我一个错误。

I know that reference is to avoid copying overhead. and const is for read-only. But here I have to make it a const reference, otherwise if I remove the "const", g++ will throw an error to me.

Edit2：我的猜测是返回对象foo（）是一个临时对象，不允许更改临时对象的值？

My guess is that the return object of foo() is a temporary object, and it's not allowed to change the value of a temporary object ?

推荐答案

错误消息，我不确定编译器可能会抱怨什么，但我可以解释逻辑上的原因：

Without the error message, I'm not exactly sure what the compiler might be complaining about, but I can explain the reason logically:

在行：

bar(foo());

foo（）的返回值是一个临时A;它是通过调用foo（）创建的，然后一旦bar（）返回就被销毁。执行非const操作（即更改临时A的操作）没有意义，因为对象A将在之后被销毁。

The return value of foo() is a temporary A; it is created by the call to foo(), and then destructed as soon as bar() returns. Performing a non-const operation (i.e. an operation that changes the temporary A) doesn't make sense, as the object A is destructed right afterwards.

多看一点，这是这个问题的虚拟重复：

Looking a little more, this is a virtual dup of this question:

http://stackoverflow.com/questions/1565600/non-const-reference-to-temporary-object

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