C++, passing struct features into function as a parameter

Question

First of all, I would like to say this is the first question I am asking on stackOverflow, so I apologize if I am not clear enough.

My question regards referring parametrically to a struct feature inside a function. I work in C++.

What I really want to achieve is to be able to sort a vector of struct objects (or class objects) based on a specific struct feature, which is given as a parameter. I also want to give the type of struct through a template, so some workarounds that deal with specific situations might not work in general.

I will show a simplistic example of what I mean.

Let's say, I have a struct called "human" with features: "age", "height", "weight".

Let's also assume that I have a vector of "human" objects called "mankind".

Here, let's say I want to make a function that can output to the screen each element's age, height or weight, depending on what I pass as a parameter.

The code below obviously doesn't work. I am asking for the proper way to do this.

struct human{
    int age;
    int height;
    int weight;
};

void show(vector<human> &elements, int value){
    for (int i=0; i<elements.size(); i++)
        cout << elements[i].value << endl;
}

int main{
    ...
    vector<human> mankind;
    ...
    show(mankind, age);
    show(mankind, height);
    show(mankind, weight);
    ...
    return 0;
}

I want to point out, that this example is a very simple case. Of course, I can make this work if I make separate functions for each feature or if I use a cheeky way, like passing a string "age" or "height" or "weight" as a parameter, checking it inside the function and having a completely separate case for each one.

However, such workarounds won't work in the general case of the problem, especially if I have many different types of structs (passed through a template T and vector< T > ) and features.

Answer 1

One way to do this is to use pointers-to-members, a C++ feature that lets you create pointers that refer to specific fields of a struct or class.

The syntax here might look a little scary, but it's actually not so bad. For example, here's how you'd get a pointer to the height field of your human struct:

int human::* ptr = &human::height;

Here, the syntax int human::* ptr means

1. that it's a pointer to something inside the human type;
2. specifically, it's a pointer to an int data member; and
3. that data member is the height field.

Once you have this pointer, you can combine it with a human struct to isolate out just that member like this:

human agatha;
cout << agatha.*ptr << endl;

Here, the .* operator means "pull up the field pointed at by ptr.

In your case, you might put things together like this:

void show(vector<human> &elements, int human::* value){
    for (int i=0; i<elements.size(); i++)
        cout << elements[i].*value << endl;
}

int main{
    ...
    vector<human> mankind;
    ...
    show(mankind, &human::age);
    show(mankind, &human::height);
    show(mankind, &human::weight);
    ...
    return 0;
}

Answer 2

I prefer the lambda approach to this problem. If the function doesn't know what will be printed, just the calling code, then you can pass a lambda to the function that contains the code of what to print. That way you can make the function completely generic like

template <typename T, typename Func>
void show(vector<T> const& elements, Func f){
    for (auto const& e : elements)
        cout << f(e)<< endl;
}

and then you would call it like

show(mankind, [](auto const& e){return e.age;});
show(mankind, [](auto const& e){return e.height;});
show(mankind, [](auto const& e){return e.weight;});

---

If show needs to show this member in order then you can even leverage the lambda in you call to std::sort like

template <typename T, typename Func>
void show(vector<T>& elements, Func f){
    std::sort(elements.begin(), elements.end(), [=](auto const& lhs, auto const& rhs){return f(lhs) < f(rhs);});
    for (auto const& e : elements)
        cout << f(e)<< endl;
}

So the element to print is used to the sort vector and to print the element.