在C ++中重载运算符时指定“ const”

Question

Code:

4: typedef unsigned short  USHORT;
5: #include <iostream.h>
6:
7:     class Counter
8:     {
9:        public:
10:          Counter();
11:          ~Counter(){}
12:          USHORT GetItsVal()const { return itsVal; }
13:          void SetItsVal(USHORT x) {itsVal = x; }
14:          void Increment() { ++itsVal; }
15:          const Counter& operator++ ();
16:
17:       private:
18:          USHORT itsVal;
19:
20:    };
21:
22:    Counter::Counter():
23:    itsVal(0)
24:    {};
25:
26:    const Counter& Counter::operator++()
27:    {
28:       ++itsVal;
29:       return *this;
30:    }
31:
32:    int main()
33:    {
34:       Counter i;
35:       cout << "The value of i is " << i.GetItsVal() << endl;
36:       i.Increment();
37:       cout << "The value of i is " << i.GetItsVal() << endl;
38:       ++i;
39:       cout << "The value of i is " << i.GetItsVal() << endl;
40:       Counter a = ++i;
41:       cout << "The value of a: " << a.GetItsVal();
42:       cout << " and i: " << i.GetItsVal() << endl;
48:     return 0;
49: }

I'm studying overloading operators in C++ and can't wrap my head around the "const" specifier in line 26. The way I understood constant reference is that we are not allowed to change the value that is stored in the reference. But inside the operator++ function (lines 26-30), the member variable "itsVal" is incremented. Doesn't this violate the "const" requirement in the function's definition?

Answer 1

The operator is returning a reference to an internal parameter as a const reference, which means that client code can not modify the reference they receive from the operator.

If, on the other hand the member function itself was const:

const Counter& Counter::operator++() const

then the function would not be allowed to modify any of its members. As it stands it can do any modification it wants before returning the reference.