C print a structure with cout-CodePudding

I have been trying to find a simple way to return 2 values from a function and found online that creating a structure to store the values was the easiest method of doing so. Now I have written the structure and the function I can not figure out how to actually print the structure. I have seen many other posts about this online, but I don't understand the answers to them. Could anybody please explain to me how I can fix my code without using any technical terms (without explanation). Preferably an addition/modification of my current code with some research material on the subject. I am not experienced in coding at all, nor do I have an education centered around coding. I am just doing this in my spare time for a bit of fun.

#include <iostream>
#include <cmath>

struct values {
    float value1;
    float value2;
};

values quadratic(int a, int b, int c) {
    float d = sqrt(b*b - 4*a*c);
    float x1 = (-b   d)/2*a;
    float x2 = (-b - d)/2*a;
    values result = {x1, x2};
    return result;
};

int main() {
    std::cout << quadratic(1, 2, -1);
};

CodePudding user response：

There are at least three ways to do this.

Method 1: Get the struct result and print it

This is the simplest solution.

int main() {
    values vs = quadratic(1, 2, -1);
    std::cout << vs.value1 << ", " << vs.value2 << "\n";
}

Method 2: Use automatic structured bindings

This requires C 17 minimum (I think, you shouldn’t be using anything less anyway):

int main() {
    auto [a, b] = quadratic(1, 2, -1);
    std::cout << a << ", " << b << "\n";
}

It’s prettier than Method 1, but basically the same thing.

Method 3: Overload the stream insertion operator

This is where you question was marked as a duplicate and linked to do do exactly this:

std::ostream & operator << (std::ostream & outs, const values & vs) {
    return outs << vs.value1 << ", " << vs.value2;
}

int main() {
  std::cout << quadratic(1, 2, -1) << "\n";
}

This is perhaps the prettiest at the call site, but does require your values struct to be convertible into a string (which is what the overloaded << operator does).

Method 4: Return a tuple instead of a custom struct

This is basically a rehash of 2.

#include <cmath>
#include <iostream>
#include <tuple>

std::tuple <float, float> quadratic(int a, int b, int c) {
    float d = sqrt(b*b - 4*a*c);
    float x1 = (-b   d)/2*a;
    float x2 = (-b - d)/2*a;
    return std::make_tuple(x1, x2);
};

int main() {
    auto [a, b] = quadratic(1, 2, -1);
    std::cout << a << ", " << b << "\n";
};

That’s all I’m gonna bother with off the top of my head, but I think it covers all the bases.

EDIT: fixed dumb thing I wrote

CodePudding user response：

One common way to do this is to overload operator<<. See this question for further information.

struct values {
    friend std::ostream& operator<<(std::ostream&, const values&);

    float value1;
    float value2;
};

std::ostream& operator<<(std::ostream& os, const values& v) {
  return os << v.value1 << ", " << v.value2;
}