Returning counters to 0 ?

[mgirl]

Hi,
My change counting program runs perfectly except the counters don't return to 0 when I try another number. The change adds up from the first try. Is there any way to fix this?

#include <stdio.h>

int main()
{
    int toonie=0, dollar=0, quarter=0, dime=0, nickel=0, cent=0;  
    float amount, change, cchange;
    char ch='Y', N, Y;
    
    while(ch=='Y'){
         
    printf("\n Your total is $", amount);
    scanf(" %f", &amount);
    
    change=(5.00-amount);
    printf("\n Your change is $%3.2f\n", change);
    while(change>=0 && change<5.00){
    cchange=(change*100);
    while (cchange>=200){
        cchange-=200;
        toonie++;
        }
    while (cchange>=100){
        cchange-=100;
        dollar++;
        }
    while (cchange>=25){
        cchange-=25;
        quarter++;
        }
    while (cchange>=10){
        cchange-=10;
        dime++;
        }
    while (cchange>=5){
        cchange-=5;
        nickel++;
        }
    while(cchange>=1){
        cchange-=1;           
        cent++;
        }
        
       // print results
       printf(" You have %d coins of 2 dollars \n", toonie);
       printf(" You have %d coins of 1 dollar \n", dollar);
       printf(" You have %d coins of 25 cents \n", quarter);
       printf(" You have %d coins of 10 cents \n", dime);
       printf(" You have %d coins of 5 cents \n", nickel);
       printf(" You have %d coins of 1 cent \n", cent);
       break;
       } // end while
       
       printf("\n Do you want to try again? (Y/N)\n\n");
       ch=toupper(getche());
       
       } // end while
       getch();
       
       }

thanks for your help!

[Dave Sinkula]

Why not put the initializations in the loop?

[mgirl]

Why not put the initializations in the loop?

I had thought of that but didn't know if it was ok to do it.
Thanks. It works!

[cikara21]

#include <stdio.h>

int main()
{
    int toonie, dollar, quarter, dime, nickel, cent;  
    float amount, change, cchange;
    char ch='Y', N, Y;
    
    while(ch=='Y'){
    //-- init / reinit
    toonie = 0; dollar = 0; quarter = 0; dime = 0; nickel = 0; cent = 0;
    printf("\n Your total is $", amount);
    scanf(" %f", &amount);
    
    change=(5.00-amount);
    printf("\n Your change is $%3.2f\n", change);
    while(change>=0 && change<5.00){
    cchange=(change*100);
    while (cchange>=200){
        cchange-=200;
        toonie++;
        }
    while (cchange>=100){
        cchange-=100;
        dollar++;
        }
    while (cchange>=25){
        cchange-=25;
        quarter++;
        }
    while (cchange>=10){
        cchange-=10;
        dime++;
        }
    while (cchange>=5){
        cchange-=5;
        nickel++;
        }
    while(cchange>=1){
        cchange-=1;           
        cent++;
        }
        
       // print results
       printf(" You have %d coins of 2 dollars \n", toonie);
       printf(" You have %d coins of 1 dollar \n", dollar);
       printf(" You have %d coins of 25 cents \n", quarter);
       printf(" You have %d coins of 10 cents \n", dime);
       printf(" You have %d coins of 5 cents \n", nickel);
       printf(" You have %d coins of 1 cent \n", cent);
       break;
       } // end while
       
       printf("\n Do you want to try again? (Y/N)\n\n");
       ch=toupper(getche());
       
       } // end while
       getch();
       
       }

[kenji]

An unwritten rule in programming, if your not sure try it anyway and at worst case it wont work it will only unleash mutually assured destruction on us.

[ajay.krish123]

your counter will set to zero when u re-execute the program.