Function access problem

raziel · #1 07-Dec-2007, 09:49

Hello everyone, I need a little help with a program, acctualy with a part of it. I'm trying to make a program that represets the functioning of a cash dispenser using functions so that I can return to the first menu at anytime, but I can't access a function that hasn't been declared up to that point.

I haven't done much with the code because of the problem above, the only compiler error is the following: "implicit declaration of function 'int main()' "

CPP / C++ / C Code:

#include<iostream>
#include<string>

int user_type=00;
int user_pass=00;
int user_option=00;

int usermenu ()
{
 do{if(user_option==00)
       {cout<<"Please select your option: "<<endl;}
    else{cout<<"Please type a valid option: "<<endl;}
    cout<<"1. Withdraw"<<endl<<"2. Change Password"<<endl;
    cin>>user_option;
/*this is the problem*/if (user_option==0) main();} while(user_option!=1 && user_option!=2 && user_option!=0);

 system("pause");
 return 0;
}

void main ()
{
 do{if(user_type==00)
      {cout<<"Welcome to the bank, please type the corresponding option for your user: "<<endl;}
    else{cout<<"Please enter a valid option: "<<endl;}
    cout<<"1. Client"<<endl<<"2. Administrator"<<endl;
    cin>>user_type;} while(user_type!=1 && user_type!=2);
    if(user_type==1)
      {do{if(user_pass==00)
         {cout<<"Please enter your user password: "<<endl;}
       else{cout<<"The password is not valid, please enter it again: "<<endl;}
       cin>>user_pass;} while(user_pass>9999 || user_pass<1000);
      }
    else if(user_type==2){}
    if(user_pass==0) main();
    else usermenu ();
}

As you can see the code is written up to the part where the user has to type one of the options provided, after that the program ends.
I want to make it work so that the user can return to the first menu typing a '0'.
Any ideas on how I could do this preferably using functions would be appreciated.

fakepoo · #2 07-Dec-2007, 10:02

I'm not sure if you can call main() like that. What I would do if I were you is take all the stuff in main() and put it into a new function called MainMenu(). This way, in main() you can just call MainMenu().

raziel · #3 07-Dec-2007, 10:26

Like this?:

CPP / C++ / C Code:

#include<iostream>
#include<string>

int user_type=00;
int user_pass=00;
int user_option=00;

int usermenu ()
{
 do{if(user_option==00)
       {cout<<"Please select your option: "<<endl;}
    else{cout<<"Please type a valid option: "<<endl;}
    cout<<"1. Withdraw"<<endl<<"2. Change Password"<<endl;
    cin>>user_option;
/*the problem still here!*/if (user_option==0) mainmenu();} while(user_option!=1 && user_option!=2 && user_option!=0);

 system("pause");
 return 0;
}

void mainmenu ()
{
 do{if(user_type==00)
      {cout<<"Welcome to the bank, please type the corresponding option for your user: "<<endl;}
    else{cout<<"Please enter a valid option: "<<endl;}
    cout<<"1. Client"<<endl<<"2. Administrator"<<endl;
    cin>>user_type;} while(user_type!=1 && user_type!=2);
    if(user_type==1)
      {do{if(user_pass==00)
         {cout<<"Please enter your user password: "<<endl;}
       else{cout<<"The password is not valid, please enter it again: "<<endl;}
       cin>>user_pass;} while(user_pass>9999 || user_pass<1000);
      }
    else if(user_type==2){}
    if(user_pass==0) mainmenu();
    else usermenu ();
}

void main ()
{
 mainmenu();
}

Please tell me if I got you wrong.
Thanks

fakepoo · #4 07-Dec-2007, 11:24

That looks like what I was saying. Also, just out of curiosity, why does usermenu() return an integer?

raziel · #5 07-Dec-2007, 15:23

Good thing I undrstood, but I keep getting almost the very same problem: implicit declaration of function 'int mainmenu ()'
It won't compile as before.

The answer for the second one is because I thought it was needed to exit the function and the program.

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