与 const 和没有相同的功能 - 何时以及为什么?

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.