具有const和不具有const的相同功能-什么时候以及为什么？ [英] Same function with const and without - When and why?

问题描述

  T& f（）{//一些代码...} 
 const T& f（）const {//一些代码...} 
  

我已经看到了几个现在（在我一直在研究的入门书中）。我知道第一个const使返回值const，换句话说：不可修改。我相信，第二个const允许也可以为const声明的变量调用该函数。

但是，为什么要在同一个类定义中同时使用两个函数呢？编译器如何区分这些？我相信第二个f（）（带有const）也可以用于非const变量。

解决方案

但是为什么要在一个类定义中同时具有相同的类定义？

同时拥有两个功能可以使您： / p>

• 在可变对象上调用该函数，并根据需要修改结果；和


• 调用 const 对象上的函数，仅查看结果。

只有第一个，您不能在 const 对象上调用它。仅使用第二个，您就无法使用它来修改它返回引用的对象。

以及编译器如何区分这些？

在调用函数时选择 const 重载 const 对象（或通过引用或指向 const 的指针）。

我相信第二个f（）（带有const）可以用于非const变量，如下所示：

如果那是唯一的重载，那就可以。在两个重载的情况下，都将选择非 const 重载。

T& f() { // some code ... }
const T& f() const { // some code ... }

I've seen this a couple of times now (in the introductory book I've been studying thus far). I know that the first const makes the return value const, in other words: unmodifiable. The second const allows that the function can be called for const declared variables as well, I believe.

But why would you have both functions in one and the same class definition? And how does the compiler distinguish between these? I believe that the second f() (with const) can be called for non-const variables as well.

解决方案

But why would you have both functions in one and the same class definition?

Having both allows you to:

• call the function on a mutable object, and modify the result if you like; and
• call the function on a const object, and only look at the result.

With only the first, you couldn't call it on a const object. With only the second, you couldn't use it to modify the object it returns a reference to.

And how does the compiler distinguish between these?

It chooses the const overload when the function is called on a const object (or via a reference or pointer to const). It chooses the other overload otherwise.

I believe that the second f() (with const) can be called for non-const variables as well.

If that were the only overload, then it could. With both overloads, the non-const overload would be selected instead.

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