Two functions variables in to a single function

Question

I have two variables in two different functions, I'd like to store them in the third function without using global variables. How to do it?

something like this

void function1() { 
  a = 1;
  b = 2;
}

void function2() {
  c = 3;
  d = 4;
}

void function3 () {
  cout << a;  
  cout << b;  
  cout << c;  
  cout << d;  
}

The best solution depends on what your code actually wants to do. With placeholder names like function1 or a, it's hard to give you an appropriate answer that does not only cover the "what's possible" aspect but also the "what's useful" aspect. — Christian Hackl
– Christian Hackl, Commented Aug 7, 2015 at 18:20
I agree with Christian. You can use a) pass-by-reference, b) return structure or class or tuple or pair c) implement as class member so on... You should choose the one that best suites your context. — pitnal
– pitnal, Commented Aug 7, 2015 at 18:28

Cory Kramer · Accepted Answer · 2015-08-07 17:54:43Z

6

Your functions can return values so you can pass variables to other functions, like so

std::pair<int, int> function1() {
    int a = 1;
    int b = 2;
    return {a, b};
}

std::pair<int, int> function2() {
    int c = 3;
    int d = 4;
    return {c, d};
}

void function3 () {
    int a, b, c, d;
    std::tie(a, b) = function1();
    std::tie(c, d) = function2();
    std::cout << a;  
    std::cout << b;  
    std::cout << c;  
    std::cout << d;  
}

Working demo

answered Aug 7, 2015 at 17:54

Cory Kramer

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5 Comments

flogram_dev Over a year ago

Or std::tuple in place of std::pair to be more generic.

ravenspoint Over a year ago

IMHO this is an awkward way to solve a simple problem. It is not at all obvious what std::tie(a, b) = function1(); does.

Cory Kramer Over a year ago

@ravenspoint .... I can't really help if people are aware of or ignorant of the standard library. If they don't know what a function does, they should look it up. Ignorance is not a reason not to use a tool for its primary purpose. Here's the documentation let the knowledge flow!

ravenspoint Over a year ago

Sorry, I disagree. making code unnecessarily complicated in order to show off your knowledge is not good practice. Self explanatory code, without reference to any documentation, is most easily maintained. Remember, debugging is harder than coding - if you write code as complicated as you know how, then you will be unable to debug it.

Cory Kramer Over a year ago

Point taken. Though I do not think that this solution is particularly difficult nor complicated, but I will leave that up to the OP. If it is understandable, straight-forwared, and solves their problem then they are free to use this solution. Otherwise they may use any other of the suggested methods.

ravenspoint · Accepted Answer · 2015-08-07 17:55:09Z

4

Make the functions methods of a class, and the variables attributes of the class.

class A
{
public:
int a;
int b;
int c;
int d;

void function1() { 
  a = 1;
  b = 2;
}

void function2() {
  c = 3;
  d = 4;
}

void function3 () {
  cout << a;  
  cout << b;  
  cout << c;  
  cout << d;  
}
};

answered Aug 7, 2015 at 17:55

ravenspoint

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3 Comments

Daniel Jour Over a year ago

While you can do this, I'd highly advise against it. This is pure side effect code.

Silvio Mayolo Over a year ago

This is only one step better than using global variables. Unless something actually fits the OO model, you shouldn't default to throwing functions in a class haphazardously.

ravenspoint Over a year ago

One step better, exactly! With minimal change to the original code, the original question is answered with straightforward syntax ( i.e no pass by reference or hiding values in pairs ) Why insist on running, when you can walk instead?

Casey · Accepted Answer · 2015-08-07 18:09:18Z

4

Use pass-by-reference:

int main() {
    int a;
    int b;
    function1(a, b);

    int c;
    int d;
    function2(c, d);

    function3(a, b, c, d);

    return 0;
}

void function1(int& a, int& b) { 
  a = 1;
  b = 2;
}

void function2(int& c, int& d) {
  c = 3;
  d = 4;
}

void function3(int a, int b, int c, int d) {
  cout << a;  
  cout << b;  
  cout << c;  
  cout << d;  
}

edited Aug 7, 2015 at 18:09

answered Aug 7, 2015 at 17:56

Casey

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3 Comments

ravenspoint Over a year ago

IMHO pass by reference should be avoided. It is hard to see that function1(a, b); actually changes the values of a,b as a side effect.

Casey Over a year ago

@ravenspoint The whole point of pass by reference is to show that the function will change the values!

ravenspoint Over a year ago

Not so. In the calling code there is no way to see whether it uses pass by value or pass by reference - the syntax is identical. You can only tell by finding the function signature.

Mr. Hello_world · Accepted Answer · 2015-08-07 18:51:15Z

2

You can pass them by reference

#include<iostream>
using namespace std;
void function1(int &a,int &b) {
    a = 1;
    b = 2;

}

void function2(int &c, int &d) {
    c = 3;
    d = 4;
}

void function3(int a1,int b1,int c1,int d1) {

    cout << a1;
    cout << b1;
    cout << c1;
    cout << d1;
}
int main(){
    int a = 0, b = 0, c = 0, d = 0;
    function1(a, b);
    function2(c, d);
    function3(a, b, c, d);



}

answered Aug 7, 2015 at 18:51

Mr. Hello_world

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