Passing arguments from function to function in c++

Question

I'm new to c++ and I'm trying to make a chess game.

Here's the thing I'm trying to make the main menu, and am having trouble. here's the snippit of my code:

int mMenu(int&, char&, bool&, char&);

int main(char&) 
{
    int choice;
    char sure;
    bool quit = false;

    char ctrl // used for the control from main menu to main()

    mMenu (choice, sure, quit);

    do
    {
        if (ctrl == a)
            NewGame();
        else if (ctrl == b)
            LoadGame();
        else
            quit = true;
    }
    while (quit == true);

    return 0;
}


int mMenu(int& choice, char& sure, bool& quit, char& ctrl,)
{
    do
{
    cout << "                           Chess                               "
         << "------------------------ Main Menu ------------------------\n"
         << "Please choose what operation you'd like to perform from the         menu below\n\n"
         << "1. New Game.\n"
         << "2. Load Game.\n"
         << "Exit.\n"
         << "Your choice: ";
    cin >> choice;

    if (choice == 1)
    {
        cout << "Are you sure you wish to start a new game? (Y/N) ";
        cin >> sure;
        if (sure != 'Y')
            clrscr();
        else
        {
            ctrl = a;
            quit = true;
    }
    else if (choice == 2)
    {
        ctrl = b;
        quit = true;
    }
    else if (choice == 3)
    {
        cout << "Are you sure you wish to exit? (Y/N) ";
        cin >> sure;
        if (sure != 'Y')
            clrscr();
        else
        {
            quit = true;
            ctrl = c;
        }
    }
    }
}
while (quit = true);

return ctrl;
}

From that code my compiler (visual c++) is saying that int the main() function, mMenu does not take 3 arguments. What is wrong and how do I make it work?

Thanks in advance.

Also as you can see I'm trying to use clrscr(); but the compiler is flagging it saying it cannot find the definition for it, despite putting in the #include any ideas?

You've defined a function that takes 4 arguments, but you've only provided 3 parameters when you try to call it. — Oliver Charlesworth
– Oliver Charlesworth, Commented Dec 4, 2011 at 0:22
int mMenu(int& choice, char& sure, bool& quit, char& ctrl = 0) last parameter with default value — Yappie
– Yappie, Commented Dec 4, 2011 at 0:28
Note: int main(char&) in the global namespace should not compile cleanly for a hosted implementation. — Cheers and hth. - Alf
– Cheers and hth. - Alf, Commented Dec 4, 2011 at 1:03
Note 2: char ctrl is missing a semicolon and shouldn't compile. — Cheers and hth. - Alf
– Cheers and hth. - Alf, Commented Dec 4, 2011 at 1:05

oh whatever · Accepted Answer · 2011-12-04 00:24:38Z

2

It really doesn't take 3 parameters, it takes 4:

int mMenu(int& choice, char& sure, bool& quit, char& ctrl,)
            // << the "," in the end 
            // shouldn't be there

How to fix it? Add the missing parameter:

mMenu (choice, sure, quit, ctrl/*<a ctrl parameter goes here>*/);

You even defined the variable ctrl, just forgot to pass it as the last argument:-)

answered Dec 4, 2011 at 0:24

oh whatever

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1 Comment

user1079559 Over a year ago

OMG! I'm such an idiot XD I guess that was THE most noobiest question you have ever answered! Thank you so much :L

parapura rajkumar · Accepted Answer · 2011-12-04 00:24:58Z

0

it is because you have defined mMenu as taking four arguments but call it only with three

mMenu (choice, sure, quit);

answered Dec 4, 2011 at 0:24

parapura rajkumar

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Comments

Daniel Daranas · Accepted Answer · 2011-12-04 00:25:58Z

0

What's wrong is that mMenu does not take 3 arguments. It takes 4.

There are two ways to make this compile:

Change mMenu so that it takes 3 arguments.
Call mMenu with its current 4 arguments.

answered Dec 4, 2011 at 0:25

Daniel Daranas

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