CODEWITHSUNDEEP
c++
sg552
sg552

Reputation: 1543

Stop send value to next function

I have this working code:

/* Include files */
#include <iostream>
#include <string>
#include <limits>

using namespace std;

void fnMainMenu(char s);

/******************************************************************************
* Function: fnUpdateSalary
* Description: Update employee salary
******************************************************************************/
void fnUpdateSalary()
{   
    int choice = 0;
    char data = 'a';
    incorrect_input: // goto teleport exit :)
    cout<<"\n\t=======================";
    cout<<"\n\tWelcome\n";
    cout<<"\t=======================\n\n";
    cout<<"1. Update employee salary\n";
    cout<<"2. Main menu\n";
    cout<<"3. Exit system\n";
    cout<<"\n >> ";

    cin>>choice;
    cin.clear();
    cin.ignore(numeric_limits<streamsize>::max(),'\n');

    switch(choice){
        case 1 : 
            //fnUpdateSalary();
            break;
        case 2 : 
            fnMainMenu(data);
            break;
        case 3 : 
            exit(1);
            break;
        default :   
        cout<<"Input not recognized. Please enter the correct input.\n";


    }
}

void fnLog()
{   
 char data = 'b';
fnMainMenu(data); // call Main Menu and I want to pass "hello"
}





/******************************************************************************
* Function: fnMainMenu
* Description: Menu for the customer 
******************************************************************************/
void fnMainMenu(char s)
{   
     cout << s;

if (s == 'a')
{ cout << "a = admin";
}
else
{cout << "\n\nb not admin";
}

    //system("cls");
    int chooice = 0;

    cout<<"\n\t=======================";
    cout<<"\n\tWelcome\n";
    cout<<"\t=======================\n\n";
    cout<<"1. Manage employee\n";
    cout<<"2. Search employee\n";
    cout<<"3. Employee report\n";
    cout<<"4. Reset password\n";
    cout<<"5. Exit system\n";
    cout<<"\n >> ";
int numbers = 2;
    cin>>chooice;
    cin.clear();
    cin.ignore(numeric_limits<streamsize>::max(),'\n');

    switch(chooice){
        case 1 : 
            fnUpdateSalary();
            break;
        case 4 : 
        //  fnChangePassword();
            break;          
        default : exit(1);

    }
}


/******************************************************************************
* Function: main
* Description: The main calling function
******************************************************************************/
int main()
{   

    fnLog();
    return 0;
}

fnLog() send b to fnMainMenu(). When I am at fnUpdateSalary() I want to go to fnMainMenu() again. My question is, is there anyway from fnUpdateSalary() or whatever function available to call fnMainMenu() without declaring variable data again? I was hoping I could just use fnMainMenu(); instead of 

char data = 'r'; fnMainMenu(data);

I get this error error C2660: 'fnMainMenu' : function does not take 0 arguments if I just called fnMainMenu();

I hope my question make sense. Thanks in advance.

Upvotes: 0

Views: 66

Answers (3)

Konrad Rudolph
Konrad Rudolph

Reputation: 545686

The parameter doesn’t seem to serve a real purpose anyway. Passing b into the main menu function certainly achieves nothing. If you just want to distinguish between admin and non-admin access, change the type and usage of the argument:

void fnMainMenu(bool is_admin) {
    if (is_admin)
        cout << "Admin\n";
    else
        cout << "Not admin\n";
}

And call it like this:

fnMainMenu(true);
// Or, alternatively:
fnMainMenu(false);

That’s it. Incidentally, you don’t need (and shouldn’t!) declare a variable to pass as the argument here. Just pass the value directly, like I’ve done above.

Also, why are your function names prefixed by fn? Don’t do this, it’s not good practice. Just use proper names that explain well what the functions do.

Upvotes: 1

Fran&#231;ois F&#233;votte
Fran&#231;ois F&#233;votte

Reputation: 20258

Since you're writing C++ code and data seems to be relatively long-lived with respect to the application logic, it would perhaps make sense to regroup these functions into a class.

data could be a data member of this class, and fnUpdateSalary, fnLog, fnMainMenu methods.

class Application {
public:
  Application ()
  : data ('b') // default value
  { }

  void fnLog() {
    data = 'b';
    fnMainMenu ();
  }

  void fnMainMenu() {
    if (data == 'a')
      cout << "a = admin";
    else
      cout << "\n\nb not admin";

    // ...
    fnUpdateSalary ();
    // ...
  }

  void fnUpdateSalary() {
    // ...
    fnMainMenu ();
    // ...
  }


private:
  char data;
};

int main () {
  Application app;
  app.fnLog ();
}

Upvotes: 0

AJ.
AJ.

Reputation: 1671

If I fully understand what you are doing, you need a combination of two things. Firstly, a static variable in fnMainMenu and secondly a default parameter:

fnMainMenu(char s = '\0')
{
  static char c;
  if(s != '\0') c = s;
  ...
  ...
}

The "static" keyword means that the character will be preserved across function calls. The default parameter means that s will be assigned a null termination character unless you explicitly pass another value.

Upvotes: 0

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