How can I make a variable always equal to the result of some calculations?

Question

In math, if z = x + y / 2, then z will always change whenever we replace the value of x and y. Can we do that in programming

Answer 1

So a big problem that I see with the lambda solutions provided is that z is calculated each time that it is inspected even if neither x nor y has changed. To get around this you really need to link these variables. I would suggest doing that via class:

class foo {
    int x;
    int y;
    int z;
    void calculate() { z = (x + y) / 2; }
    friend istream& operator >>(istream& lhs, foo& rhs);
public:
    void set_x(const int param) {
        x = param;
        calculate();
    }
    int get_x() const { return x; }
    void set_y(const int param) {
        y = param;
        calculate();
    }
    int get_y() const { return y; }
    int get_z() const { return z; }
};

istream& operator >>(istream& lhs, foo& rhs) {
    lhs >> rhs.x >> rhs.y;
    rhs.calculate();
    return lhs;
}

This will recalculate z each time x or y is set. This is a good solution if you access z frequently, and x and y are set infrequently. If x and y are set frequently or calculate is expensive you might consider:

class foo {
    int x;
    int y;
    int z;
    bool dirty;
    void calculate() { z = (x + y) / 2; }
    friend istream& operator >>(istream& lhs, foo& rhs);
public:
    void set_x(const int param) {
        x = param;
        dirty = true;
    }
    int get_x() const { return x; }
    void set_y(const int param) {
        y = param;
        dirty = true;
    }
    int get_y() const { return y; }
    int get_z() const { 
        if(dirty) {
            calculate();
        }
        return z;
    }
};

istream& operator >>(istream& lhs, foo& rhs) {
    lhs >> rhs.x >> rhs.y;
    rhs.dirty = true;
    return lhs;
}

Note that I've included an extraction operator, so whichever you choose your code can turn into something as simple as:

foo xyz;

cin >> xyz;
cout << xyz.get_z();