When I use only sort() without arguments, it returns correctly in alphabetical order. When I try to add in arguments like below, it just returns the word in the same order the string was entered in. I'm not entirely sure what I'm doing incorrectly.
var a = str.split("")
return a.sort((a,b) => a-b).join("");
Try using localeCompare:
var a = str.split("")
return a.sort((a, b) => a.localeCompare(b)).join("");
When comparing strings a - b won't work. You can use the ternary logic operator here: a < b ? -1 : 1
Place that expression in the function you pass to sort and that should do the trick.
a < b ? -1 : 1 simply returns -1 iff a < b and 1 iff a >= b. The sort method expects a number to be returned: if negative, a goes before b, if positive a goes after b. But this conditional expression isn’t semantically perfect, since it’s missing the 0 case, when a and b are equal. That’s why you should really use a.localeCompare(b) instead.Try this:
var str = 'zyxw'
var a = str.split("")
console.log(a)
var res = a.sort((a,b) => a.localeCompare(b)).join("");
console.log(res)
a - bisNaNwhenaandbare non-numeric strings. What are you trying to achieve?return a.sort((a,b) => a > b ? 1 : -1).join(""); should do itvar array = ['c','d','a']thenarray.sort(), it returns['a',''b','c']...i was wondering why when adding the arguments, it doesn't achieve the same thing