const函数可以在本地对象上调用非const函数吗？

Question

I have a question regarding const functions: can a const function call non-const functions on local objects in the function? Here's an example of what I'm talking about:

template <class T>
Set<T> Set<T>::setUnion (const Set<T> & other) const
{
    Set<T> unionSet; //create union set
    unionSet.merge(internalStorage); //merge it with our current set
    unionSet.merge(other.internalStorage); //merge it with other set, duplicates will not be added

    return unionSet;
}

This function takes two sets and returns the union of the sets. The problem though is that the merge function is not const and the merge function also calls the add function which is also not const, so I cannot see any other way to create a union set with solely these two functions given that the setUnion function has to be const.

PS: I know the solution without doing a const function, the reason I'm doing it this way is because my professor defined it as such. Thanks.

Answer 1

Yes. The only restriction on a const member function is that it can't modify the object its called on; it can modify any other (non-const) object.

Answer 2

A const member function means that the this it receives has type roughly equivalent to T const *const this ¹ .

That means you can't invoke any non-const functions via this , but it does not affect anything you do that doesn't go through this (explicitly or implicitly).

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^{1. Technically, you can't really write out the type of this truly correctly, but the type I've given here is close enough for the current discussion.}