share variable between functions

Question

#include <iostream>

using namespace std;

void lemo(){
    int shamble;
}

int main(){
    lemo();
    shamble= 2;
    cout<<shamble;
    return 0;
}

All I want is call a variable from other function. After executing console will say:

'shamble' was not declared in this scope

What can I do for this problem?

Use variable and list of objects in other files and functions.

Put them at the top of the code to make them global (outside of the functions) — 11_22_33, Mar 19 '23 at 08:00
https://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list#388282 — Jesper Juhl, Mar 19 '23 at 08:28
Does this answer your question? [The Definitive C++ Book Guide and List](https://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list) — Marcus Müller, Mar 19 '23 at 08:32

score 6 · Accepted Answer · answered Mar 19 '23 at 08:04

Your compiler is correct. shamble was not declared within the scope of main. As it stands your lemo function does nothing. It declares an unused local variable and then exits.

You could just as well write the following, and it would actually compile.

#include <iostream>

int main() {
    int shamble = 2;
    std::cout << shamble;
    return 0;
}

If you want a variable to be available to multiple functions, it can be declared "globally" but don't. This rapidly leads to code that is difficult to reason about and troubleshoot.

Rather you should pass information from one function to another via arguments and return values. Or you may wish to encapsulate that persistent state into objects.

score 2 · Answer 2 · answered Mar 19 '23 at 08:13

You can move the declaration of the shamble variable into main(), and then pass it into lemo() for it to use, eg:

#include <iostream>

using namespace std;

void lemo(int &shamble){
    shamble = 2;
}

int main(){
    int shamble;
    lemo(shamble);
    cout << shamble;
    return 0;
}

share variable between functions

2 Answers2