问题描述
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.
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
constobject, 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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