Is there a way I can make a function a parameter?

Question

I'm new to programming and I have been messing around with a bit of C++. I have a simple program that works, but I have a lot of similar functions that I want to get rid of. I figured I would just use a function for it and add parameters that I can change, but in an if-statement I have the function call itself back so it could run the same function again.

So my question trying to find a work around so I can simplify the similar functions. In the code I have provided there are only a few differences between function 1 and 2. In my actual program I have about 8 of these similar functions.

int a;
// function 1
int ba = 0;
int bb = 0;
int bc;
// function 2
int ca = 0;
int cb = 0;
int cc;

void function2() {
    cin >> a;

    if(a == 1) {
        ca = ca + 10000;
        cb = cb + 100;
        function2();
    }

    if(a == 0) {
        cb = cb + 100;
        function2();    
    } else {
        cc = ca / cb;
    }
}

void function1() {
    cin >> a;

    if(a == 1) {
        ba = ba + 10000;
        bb = bb + 100;
        function1();
    }

    if(a == 0) {
        bb = bb + 100;
        function1();

    } else {
        bc = ba / bb;
    }
}
int main() {
    function1();
    function2();
}

why not pass a parameter (an int) and then based on that increment the corresponding variables — depperm, Jul 26 '19 at 18:23
Check out [this](https://en.cppreference.com/w/cpp/language/lambda) and [this](https://stackoverflow.com/questions/40844622/use-a-lambda-as-a-parameter-for-a-c-function). Another tip, though: don't declare non-const variables outside of your functions. Functions should have *explicit* inputs and *explicit* outputs. — hegel5000, Jul 26 '19 at 18:25

Jesper Juhl · Accepted Answer · 2019-07-26T18:29:07.920

2

The easiest way would probably be to use a std::function as an argument. But you could also use a plain function pointer or a pointer to member function if that's a better match for your use case.

edited Jul 26 '19 at 18:29

answered Jul 26 '19 at 18:25

Jesper Juhl

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Igor G · Answer 2 · 2019-07-26T19:14:57.710

The example you've posted doesn't require any function parameter...

First, you'd better use a loop instead of a tail recursion. It would better convey the idea of repeating some action until certain condition is met.

Second, if all your functions are doing the same operation on different sets of data, then you'd better convert them to a class with a method:

struct MyClass
{
    void function()
    {
        for (;;)
        {
            int    a;
            std::cin >> a;

            switch (a)
            {
                case 1:
                    m_a = m_a + 10000;
                    m_b = m_b + 100;
                    continue;
                case 0:
                    m_b = m_b + 100;
                    continue;
            }
            m_c = m_a / m_b;
            break;
        }
    }

    int    m_a = 0;    // Requires C++11
    int    m_b = 0;
    int    m_c = 0;
};

MyClass    b1;
MyClass    b2;
// ...
MyClass    b8;

int main()
{
    b1.function();
    b2.function();

    // Use `b1.m_c` through `b8.m_c` here

    return 0;
}

Third, if you don't need those m_a, m_b and m_c values after the functions have completed their work, you may convert m_a and m_b into local variables, convert m_c into a function return value, and get rid of the class:

int function()
{
    int     a = 0;
    int     b = 0;
    for (;;)
    {
        int    x;
        std::cin >> x;

        switch (x)
        {
            case 1:
                a = a + 10000;
                b = b + 100;
                continue;
            case 0:
                b = b + 100;
                continue;
            default:
                return a / b;
        }
    }
}

int main()
{
    int    c1 = function();
    int    c2 = function();
    // ....
    int    c8 = function();

    // Use `c1` through `c8` here

    return 0;
}

Is there a way I can make a function a parameter?

2 Answers2