Simple "cout " i cant fix :( help plz

[BleedingHandz]

ok.. This a short menu im creating... it seem to be working prefectly fine other than the fact that it isnt showing my cout command i think and i dont know why ... it probably reeallly simple.. could u lead me in the right path

#include <stdlib.h>
#include <iostream>

using namespace std;

int order;
float billamt;          // short for bill's amount
int f;                  // varabile to stop loop
float Vat;              // to calculate vat
float total;            // to add vat to bill amount or subtotal

int menuz ()              //function named "menuz"
{
    cout << "Welcome To Evol Restaurant & Bar" <<endl;
    cout << "Please Select an Option" <<endl <<endl;
    cout << "\t EVOL's MENU" <<endl;
    cout << "1. Burger with Fries...            $13.00" <<endl;
    cout << "2. Burger with Jacket Potatoes...  $15.00" <<endl;
    cout << "3. Spagettii..                     $11.00" <<endl;
    cout << "4. Chow Mein with Pop Chicken..    $14.00" <<endl;
    cout << "5. Kid'z Hot-Dogs...               $8.00"  <<endl;
}

int main()
{
  menuz ();              //calling function "menuz"
 
    do 
{      cin >> order;
           if (order == 1)                      // when user press 1 
              billamt += 13.00;                 // billamt = billamt + 13.00
           if (order == 2)
              billamt += 15.00;
           if (order == 3)
              billamt += 11.00;
           if (order == 4)
              billamt += 14.00;
           if (order == 5)
              billamt += 8.00;
           
}          while (order != f);    // f is the letter i want to stop tbe loop with 
           
            Vat = billamt * 0.175;
            total = billamt + Vat;
  
  cout << "Your Current Subtotal is... $" << billamt;   // here my problem i assume the program is working but this isnt beening excuted
  cout << "Your Final Bill is... $" <<total <<endl;     // this is also not being excuted
            
system ("PAUSE");
return (0);
}

[Lynx876]

int menuz() int. Assumes you'll be returning and int.

Try changing this to void, as you don't need any values to return.

[BleedingHandz]

can you simplify that for me please... Do I put void command somewhere are something im a bit lost... since im not familar with the void command

[Lynx876]

void menuz()
{
     //code...
}

void, simply means you wont be returning value;

Whereas:

int main()
{
    //code...

    return 0;
}

int main() returns 0( an int ).

[Lynx876]

Sorry, I see your problem now!

You never give int f a value!

Maybe;

int f = 0;

int menuz ()              //function named "menuz"
{
    //code...
    cout << "Enter 0 to finish order." <<endl;
}

Then in main, add a check to see if the user wants to end the order. Making order == f.

[mirec]

I am also new here but all I think is that your code is ok.. exept you can change int to void but that should not be problem to execute last two cout... I think you are pressing f after you are done billing and you used integer for f so if you press 0 or other key it should work as you want. is that your problem?? and i would also use switch statment rather than so many if but it is your program and it works like this fine.. i hope i helped...

[Lynx876]

On a side note. You're declaring your variables in the global scope.

As you only use these within the main() function, declare these within main();

int main()
{
    int order;
    float billamt;          // short for bill's amount
    int f;                  // varabile to stop loop
    float Vat;              // to calculate vat
    float total;            // to add vat to bill amount or subtotal

    //code...

    return 0;
}

By declaring these outside of main(), other functions can change the values! This could result in unwanted behavior.

Also. If you're going to set a value that cannot be changed, i.e., your value of f, which is the way to exit the loop.

Declare this as const int f = 0;. If something were to change this value, then your program will loop indefinitely.

[hayfrend]

I ran your code and figured out your problem.

first you need to set your int variable f to a default value and make it constant just in case:

const int f = 0;

THEN you need to change your menuz function to void as it doesn't return a value:

void menuz()
{}

FINALLY (and this is the most important) you need to change your while statement to this:

while(order == f);

the reason why you don't want !=(not equals) is because you are telling the program to loop as long as order does not equal 0 (what we set f to). When the user types in a 1, 2, 3, 4, or 5 then the variable order changes to either 1, 2, 3, 4, or 5 thus making order != f, so the loop continues.

Hope that helps you

[BleedingHandz]

Thankz
Mirec, - Thankz for the input :3

Lynx876, - You Showed and Taught me alot dude thankz a grand!! :I

Hayfrend. - Umm the while (order ==f); u were trying to show me wasnt working but u helped me realise where i was going wrong with my problem with your explaination so thank ... cause the value of order would change ever time the user enter a number so thank for pointing that out for me.

Thankz again guyz

[Lynx876]

No problem, fella.

And hayfrend! If he did:
while order == f );

That would mean that the user would HAVE to enter the 'exit code' to keep running the program. Seeing as 'f' if the exit from the program.