Infinite loop although I put a break;

Question

The aim of this algorithm is to return an organized array (with currencies) of the change .

I built a nested while loop to loop as long as "change" is not equal 0 then each loop in the while loop function while "change" is more than 0 (i.e : not -ve).

What have I missed?

// function checkCashRegister(price, cash, cid) { var change = cash-price; var chArr = [["PENNY", 0], ["NICKEL", 0], ["DIME", 0], ["QUARTER", 0], ["ONE", 0], ["FIVE", 0], ["TEN", 0], ["TWENTY", 0], ["ONE HUNDRED", 0]]; while (change !== 0) { while (change - 100 > 0) { chArr[8][1] += 100; change -=100; if (change <= 0) {break;} } while (change - 20 > 0) { chArr[7][1] += 20; change -=20; if (change <= 0) {break;} } while (change - 10 > 0) { chArr[6][1] += 10; change -=10; if (change <= 0) {break;} } while (change - 5 > 0) { chArr[5][1] += 5; change -=5; if (change <= 0) {break;} } while (change - 1 > 0) { chArr[4][1] += 1; change -=1; if (change <= 0) {break;} } while (change - 0.25 > 0) { chArr[3][1] += 0.25; change -=0.25; if (change <= 0) {break;} } while (change - 0.1 > 0) { chArr[2][1] += 0.1; change -=0.1; if (change <= 0) {break;} } while (change - 0.05 > 0) { chArr[1][1] += 0.05; change -=0.05; if (change <= 0) {break;} } while (change - 0.01 > 0) { chArr[0][1] += 0.01; change -=0.01; if (change <= 0) {break;} } if (change <= 0) {break;} } // Here is your change, ma'am. return chArr; } checkCashRegister(17.46, 20.00, [["PENNY", 1.01], ["NICKEL", 2.05], ["DIME", 3.10], ["QUARTER", 4.25], ["ONE", 90.00], ["FIVE", 55.00], ["TEN", 20.00], ["TWENTY", 60.00], ["ONE HUNDRED", 100.00]]);

@Thomas There's a break in the outer loop too. But yeah it's overly complex — Hespen
– Hespen, Commented Nov 17, 2016 at 9:47
Try to go through it step by step with a browser's developer tools? — RemcoGerlich
– RemcoGerlich, Commented Nov 17, 2016 at 9:47
@BahaaZidan Please use modulo... This is a first step to simplify your code. Seriously. — Amessihel
– Amessihel, Commented Nov 17, 2016 at 9:48

Nina Scholz · Accepted Answer · 2016-11-17 10:38:25Z

You need just to check the single cash parts and use there only the check if the rest is greater then the change. No need for a breaking condition inside of the while loop.

What is missing, is to check if the change is sufficient.

function checkCashRegister(price, cash, cid) { var change = cash-price; var chArr = [["PENNY", 0], ["NICKEL", 0], ["DIME", 0], ["QUARTER", 0], ["ONE", 0], ["FIVE", 0], ["TEN", 0], ["TWENTY", 0], ["ONE HUNDRED", 0]]; while (change >= 100 && cid[8][1] >= 100) { chArr[8][1] += 100; cid[8][1] -= 100; change -=100; } while (change >= 20 && cid[7][1] >= 20) { chArr[7][1] += 20; cid[7][1] -= 20; change -=20; } while (change >= 10 && cid[6][1] >= 10) { chArr[6][1] += 10; cid[6][1] -= 10; change -=10; } while (change >= 5 && cid[5][1] >= 5) { chArr[5][1] += 5; cid[5][1] -= 5; change -=5; } while (change >= 1 && cid[4][1] >= 1) { chArr[4][1] += 1; cid[4][1] -= 1; change -=1; } while (change >= 0.25 && cid[3][1] >= 0.25) { chArr[3][1] += 0.25; cid[3][1] -= 0.25; change -=0.25; } while (change >= 0.1 && cid[2][1] >= 0.1) { chArr[2][1] += 0.1; cid[2][1] -= 0.1; change -=0.1; } while (change >= 0.05 && cid[1][1] >= 0.05) { chArr[1][1] += 0.05; cid[1][1] -= 0.05; change -=0.05; } while (change > 0 && cid[0][1] >= 0.01) { chArr[0][1] += 0.01; cid[0][1] -= 0.01; change -=0.01; } // Here is your change, ma'am. return chArr; } var register1 = [["PENNY", 1.01], ["NICKEL", 2.05], ["DIME", 3.10], ["QUARTER", 4.25], ["ONE", 90.00], ["FIVE", 55.00], ["TEN", 20.00], ["TWENTY", 60.00], ["ONE HUNDRED", 100.00]], register2 = [["PENNY", 1.01], ["NICKEL", 2.05], ["DIME", 3.10], ["QUARTER", 4.25], ["ONE", 90.00], ["FIVE", 55.00], ["TEN", 20.00], ["TWENTY", 60.00], ["ONE HUNDRED", 100.00]]; console.log(checkCashRegister(17.46, 20.00, register1)); console.log(register1); console.log(checkCashRegister(19.50, 20.00, register2)); console.log(register2);

.as-console-wrapper { max-height: 100% !important; top: 0; }

And of course, you could put the coin values in an array too and make a for loop around a single while; but let's not do that work for him :-)
you could streamline it, and i could do it, but i think, the op is learning, first to deal with while loops.
i have a decimal error :\ it should return [["PENNY", 0], ["NICKEL", 0], ["DIME", 0], ["QUARTER", 0.50], ["ONE", 0], ["FIVE", 0], ["TEN", 0], ["TWENTY", 0], ["ONE HUNDRED", 0]] but it's returning [["PENNY", 0.04], ["NICKEL", 0], ["DIME", 0.2], ["QUARTER", 0.25], ["ONE", 0], ["FIVE", 0], ["TEN", 0], ["TWENTY", 0], ["ONE HUNDRED", 0]]

Amessihel · Accepted Answer · 2016-11-17 10:39:11Z

I think you should rethink your code :

function checkCashRegister(price, cash) { var change = cash - price; var chObj = { PENNY: 0, NICKEL: 0, QUARTER: 0, ONE: 0, FIVE: 0, TEN: 0, TWENTY: 0, ONE_HUNDRED: 0 } // Performs an euclidian division to get the quotient // (in 750$, there are 7 times 100 ie quotient = 7) chObj.ONE_HUNDRED = Math.floor(change / 100); // Now get the remainder (750 minus 7 times 100) change = change % 100; // So on... chObj.TWENTY = Math.floor(change / 20); change = change % 20; chObj.TEN = Math.floor(change / 10); change = change % 10; chObj.FIVE = Math.floor(change / 5); change = change % 5; chObj.ONE = Math.floor(change); change = Math.floor((change - Math.floor(change)) * 100); chObj.QUARTER = Math.floor(change / 20) / 100; change = change % 25; chObj.NICKEL = Math.floor(change / 10) / 100; change = change % 10; chObj.NICKEL = Math.floor(change) / 100; return chObj; } var chObj = checkCashRegister(245, 1000); for (var key in chObj) { if (chObj.hasOwnProperty(key)) { console.log(key + " : \t" + chObj[key]); } }

Result (viewable through the console) :

PENNY : 0 NICKEL : 0 QUARTER : 0 ONE : 0 FIVE : 1 TEN : 1 TWENTY : 2 ONE_HUNDRED : 7

It can be more improved i think. See the fiddle here.

This one assumes the cash register has unlimited amounts of each denomination
Yes, but the intention is clear - still kudos for using modulo
In that case I think that cash variable is redundant. We only need denomination (plus : there a risk that cash won't match the sum of all denominations). Thanks :)
In this case, this code will become a mess if not more optimized.

Community · Accepted Answer · 2017-05-23 10:30:28Z

your issue is that you're checking equality to 0 and your actual value is 0.00..[something] you can see that it's a known issue here

change the last if statement to be if (change.toFixed(2) <= 0) {break;}

BTW I suggest you to change all your money calculations to be on cents in order to avoid this kind of issues (instead of 1.0$ use 100 cents).

RemcoGerlich · Accepted Answer · 2016-11-17 09:57:47Z

Here is the reason: the final coin of each kind is never added to the array.

Say the amount is 0.02, 2 cents. Then in the final loop,

while (change - 0.01 > 0) { chArr[0][1] += 0.01; change -=0.01; if (change <= 0) {break;} }

The guard (0.02 - 0.01 > 0) is true, so the loop is entered and change becomes 0.01.

Next time, 0.01 - 0.01 > 0 is false, so the loop is not entered.

And then the outer loop runs again and again and again, because change is still 0.01 and it never becomes lower.

So that > should have been >=. Then all the breaks as well as the whole outer loop can be removed.

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