Comments on: Code Challenge: Change Dispenser http://blog.stevenlevithan.com/archives/change-dispenser A JavaScript and regular expression centric blog Thu, 02 Feb 2012 02:09:46 +0000 hourly 1 http://wordpress.org/?v=3.0.1 By: lawrence http://blog.stevenlevithan.com/archives/change-dispenser/comment-page-1#comment-163232 lawrence Tue, 13 Dec 2011 11:59:35 +0000 http://blog.stevenlevithan.com/?p=79#comment-163232 i used the top method and it worked. thank you, i have been looking for this html code for a while. i used the top method and it worked. thank you, i have been looking for this html code for a while.

]]> By: NEX-C3 http://blog.stevenlevithan.com/archives/change-dispenser/comment-page-1#comment-155762 NEX-C3 Thu, 17 Nov 2011 23:22:00 +0000 http://blog.stevenlevithan.com/?p=79#comment-155762 I’m impressed!!! Really informative blog post here my friend. I just wanted to comment u0026 say keep up the quality work. I’m impressed!!! Really informative blog post here my friend. I just wanted to comment u0026 say keep up the quality work.

]]> By: Prestaul http://blog.stevenlevithan.com/archives/change-dispenser/comment-page-1#comment-85164 Prestaul Tue, 08 Mar 2011 22:24:11 +0000 http://blog.stevenlevithan.com/?p=79#comment-85164 Does shorter and no "var" statements == better? Probably not, but it was fun to slap together and I enjoy taking advantage of array methods whenever possible. function makeChange(money) { return [ [100, 'dollar'], [25, 'quarter'], [10, 'dime'], [5, 'nickel'], [1, 'penny', 'pennies'] ].reduce(function(state, coin, cnt) { if(cnt = (state.amount / coin[0]) ^ 0) { state.change.push(cnt + ' ' + (cnt > 1 ? coin[2] || coin[1] + 's' : coin[1])); state.amount %= coin[0]; } return state; }, { amount: money * 100, change: [] }).change.join(', '); } Does shorter and no “var” statements == better? Probably not, but it was fun to slap together and I enjoy taking advantage of array methods whenever possible.

function makeChange(money) {
return [
[100, 'dollar'],
[25, 'quarter'],
[10, 'dime'],
[5, 'nickel'],
[1, 'penny', 'pennies']
].reduce(function(state, coin, cnt) {
if(cnt = (state.amount / coin[0]) ^ 0) {
state.change.push(cnt + ‘ ‘ + (cnt > 1 ? coin[2] || coin[1] + ‘s’ : coin[1]));
state.amount %= coin[0];
}
return state;
}, { amount: money * 100, change: [] }).change.join(‘, ‘);
}

]]> By: John Vivenzio http://blog.stevenlevithan.com/archives/change-dispenser/comment-page-1#comment-32071 John Vivenzio Sat, 02 May 2009 06:27:15 +0000 http://blog.stevenlevithan.com/?p=79#comment-32071 Obscure bug? When testing just now I found if you enter makeChange(279.96 * 100) - it rounds to 27995.9999 when multiplied by 100 - very bizarre. ...using Firefox 3.5b4 - reminds me of the Pentium 59.999... days. Obscure bug?

When testing just now I found if you enter makeChange(279.96 * 100) – it rounds to 27995.9999 when multiplied by 100 – very bizarre. …using Firefox 3.5b4 – reminds me of the Pentium 59.999… days.

]]> By: John Vivenzio http://blog.stevenlevithan.com/archives/change-dispenser/comment-page-1#comment-32070 John Vivenzio Sat, 02 May 2009 05:58:45 +0000 http://blog.stevenlevithan.com/?p=79#comment-32070 Last One... with inventory of bills and coins and shows remainder if can't make exact change. var b = [ // checkInventory to null out what's not available (on hand is b[i][1]) [10000, 0, "Hundred"], [ 5000, 1, "Fifty", "Fifties"], [ 2000, 0, "Twenty", "Twenties"], [ 1000, 0, "Ten"], [ 500, 0, "Five"], [ 200, 0, "Two"], [ 100, 0, "Dollar"], [ 25, 0, "quarter"], [ 10, 0, "dime"], [ 5, 0, "nickel"], [ 1, 0, "penny", "pennies"], [ 0] ]; function makeChange(money) { if (b[i][1] && (x = Math.floor(money / b[i][0]))) output.push(x + " " + (x <= 1 ? b[i][2] : !b[i][3] ? b[i][2] + "s" : b[i][3])); return isNaN(b[i + 1]) ? makeChange(b[i][1] ? money % b[i++][0] : (i++, money)) : output.push("Remainder: $" + money / 100), output; } makeChange(123279.96 * 100).join("\n"); Last One… with inventory of bills and coins and shows remainder if can’t make exact change.

var b = [
// checkInventory to null out what's not available (on hand is b[i][1])
[10000, 0, "Hundred"],
[ 5000, 1, "Fifty", "Fifties"],
[ 2000, 0, "Twenty", "Twenties"],
[ 1000, 0, "Ten"],
[ 500, 0, "Five"],
[ 200, 0, "Two"],
[ 100, 0, "Dollar"],
[ 25, 0, "quarter"],
[ 10, 0, "dime"],
[ 5, 0, "nickel"],
[ 1, 0, "penny", "pennies"],
[ 0]
];

function makeChange(money) {
if (b[i][1] && (x = Math.floor(money / b[i][0])))
output.push(x + ” ” + (x <= 1 ? b[i][2] : !b[i][3] ? b[i][2] + “s” : b[i][3]));
return isNaN(b[i + 1]) ? makeChange(b[i][1] ? money % b[i++][0] : (i++, money)) :
output.push(“Remainder: $” + money / 100), output;
}

makeChange(123279.96 * 100).join(“\n”);

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