Cash change program using loops and if/else

Question

Coding a program that allows user to input a cash amount up to $200.00 and then computes and prints its value in the following denominations (20, 10, 5, 1, .25, .10, .05, .0

Answer 1

Try this... Tested and it works. You problem was in the way reminder was being calculated. You need to divide the amt_ent again with previous step. Again all % operations work only with integers. So you need to convert your double to integer domain by multiplying with 100 before proceeding with calculations.

#include<stdio.h>
#include<stdlib.h>
main()
{
double amt_ent1;
int amt_ent;
int twenty, ten, five, one, quarter, dime, nickel, penny;
do
{ 
    printf ("Enter a dollar amount up to $200.00:");   //<== it is put in due to get the statement again else it execute with the same value.
    scanf ("%lf", &amt_ent1);
    amt_ent = (amt_ent1*100)/100;
    printf ("Name - Assignment 2 - Change-O-Matic\n");
    printf ("Amount entered: $%.2lf\n", amt_ent1);
    printf ("Change breakdown:\n");

    if ((amt_ent > 200.00) || (amt_ent < 00.00)) continue;

    amt_ent = amt_ent1 * 100;
    { /*Change in twenties*/
    twenty= (int) amt_ent/2000;
    if (twenty >= 2)
        printf("%i\t$20.00s\n", twenty);
    if (twenty == 1)
        printf ("%i\t$20.00\n", twenty);

        /*Change in tens*/
    amt_ent = amt_ent % 2000;
    ten = amt_ent/1000;


    if (ten >=2)
        printf ("%i\t$10.00s\n", ten);
    if (ten == 1)
        printf ("%i\t$10.00\n", ten);

        /*Change in fives*/
    amt_ent = amt_ent % 1000;
    five = amt_ent/500;

    if (five >= 2)
        printf ("%i\t$5.00s\n", five);
    if (five == 1)
        printf ("%i\t$5.00\n", five);

        /*Change in ones*/
    amt_ent = amt_ent % 500;
    one = amt_ent/100;

    if (one >= 2)
        printf ("%i\t$1.00s\n", one);
    if (one == 1)
        printf ("%i\t$1.00\n", one);

        /*Change in quarters*/
    amt_ent = amt_ent % 100;
    quarter = amt_ent/25;

    if (quarter >= 2)
        printf ("%i\t$.25s\n", quarter);
    if (quarter == 1)
        printf ("%i\t$.25\n", quarter);

    /*Change in dimes*/
    amt_ent = amt_ent % 25;
    dime = amt_ent/10;

    if (dime >= 2)
        printf ("%i\t$.10s\n", dime);
    if (dime == 1)
        printf ("%i\t$.10\n", dime);

        /*Change in nickels*/
    amt_ent = amt_ent % 10;
    nickel = amt_ent/5;

    if (nickel >= 2)
        printf ("%i\t$.05s\n", nickel);
    if (nickel == 1)
        printf ("%i\t$.05\n", nickel);

        /*Change in pennies*/
    amt_ent = amt_ent % 5;
    penny = amt_ent/1;

    if (penny >= 2)
        printf ("%i\t$.01s\n", penny);
    if (penny == 1)
        printf ("%i\t$.01\n", penny);
    }
}

while ((amt_ent <= 2000) && (amt_ent >= 0));  //<== it is put inside the do loop it is wrong, it come outside the do lopp.

return 0;
}

Answer 2

As regarding you error, there is extra brace present after printf ("Change breakdown:\n");. And you need not to put closing brace after while statement. while ((amt_ent <= 200.00) && (amt_ent >= 00.00));} Remove that also.

For you valid amount handling problem, there is a continue command which skips the remaining loop and reiterate from beginning when encountered. You can use that.

do
  { 
      printf ("Enter a dollar amount up to $200.00:");
      scanf ("%lf", &amt_ent);
      if(amount_ent < 00.00 || amount_ent>200.00)
           continue;
      printf ("Name - Assignment 2 - Change-O-Matic\n");
      printf ("Amount entered: $%.2lf\n", ((amt_ent*100)/100));
      printf ("Change breakdown:\n");

If amount entered is invalid, the continue statement will execute and remains loop is skipped. Then printf ("Enter a dollar amount up to $200.00:"); will be executed. So you can see, user wont be able to go ahead of continue unless he enters correct value of amount.