我如何从无效中脱离当前范围，以便回到主要领域？ 我需要做什么？

Question

My error:

|21|error: 'main' was not declared in this scope|
||=== Build failed: 1 error(s), 0 warning(s) (0 minute(s), 0 second(s)) ===|

My actual program is radically different and much larger (hence the unneeded headers). I just wrote this to demonstrate my issue quickly. How do I get out of the scope of that void function and get back to main (where the menu of my program actually lies)?

#include <iostream>
#include <cmath>
#include <cstdlib>
#include <cstdio>
#include <limits>

int continueint;

using namespace std;

class handles {
public:
void continueChoice()
{
    cout<<"[1] Go back to the main menu"<<endl;
    cout<<"[2] Terminate the program"<<endl;
    cin>>continueint;
    switch (continueint)
    {
    case 1:
        main();   // This is where my problem lies
    case 2:
        break;
    }
}
}handlers;

int main(int nNumberofArgs, char*pszArgs[])
{
cout<<"A message here."<<endl;
system("PAUSE");
handlers.continueChoice(); //Calls the void function in the above class
}

Answer 1

Just return:

void somefunc() {
    if (something)
        if (something_else)
            return;
        }
    }
}

You can return from a void function. You just cant return an object.

As a further note, you should never be calling main . You could make this compile by putting continueChoice under main and forward declaring it, but you would then be creating an infinite recursive loop which is very bad for everything.