Sort an Array Alphabetically using the sort Method-let a,b;

JavaScript
1

What will happen if we define variables let a,b then call arr.sort(a,b => a > b)?
I have done both ways but there was something wrong with this way… what goes behind the scene I coouldn’t get…!?

Your code so far


function alphabeticalOrder(arr) {
  // Add your code below this line
  let a,b;
  return arr;
  
  // Add your code above this line
}
console.log(alphabeticalOrder([91,89,3, 5, 1, 0, 9, 2]));

Your browser information:

User Agent is: Mozilla/5.0 (Windows NT 10.0; Win64; x64; rv:65.0) Gecko/20100101 Firefox/65.0.

Link to the challenge:
https://learn.freecodecamp.org/javascript-algorithms-and-data-structures/functional-programming/sort-an-array-alphabetically-using-the-sort-method

2

Can you give an example of something going wrong?

Outside variables that are named the same won’t matter because Array.sort() takes in a callback function not any sorts of parameters. Like this.

var numbers = [4, 2, 5, 1, 3];
numbers.sort(function(a, b) {
  return a - b;
});
console.log(numbers);

// [1, 2, 3, 4, 5]
3

You are right if each number in the array is single but try the way I’ve done. it sorts 89 before 9

4

Sort expect a positive, negative or 0 value - with a comparison you are returning negative (false) or positive (true) values, and never negative, as such you will get unexpected results

Read the documentation on sort linked above

5

Sorry to say that, I think You haven’t got what the question is about?
It is correct but I wanted to test it with separate defined variables let a; let b;
If we do it by parameters a, b it works but not with separate defined variables.

6

You have not showed your code with that, you have just written let a,b; return arr; which has nothing to do with sort

Could you post the code you mean?

You also need to consider this

7

Refer to first line of the question, please

8

Refer to the piece of post I have quoted above

9

I understand, but that is different
let’s see the combined code

function alphabeticalOrder(arr) {
  // Add your code below this line
  let a,b;
  return arr.sort(a,b => a > b);
  
  // Add your code above this line
}
console.log(alphabeticalOrder([91,89,3, 5, 1, 0, 9, 2]));

// 0,1,2,3,5,89,9,91
see the output

try the code yourself and notice the numbers 89 vs 9

10

The error is inside the sort method
Your callback never returns a negative number, so it will never sort correctly

If you use the following you don’t get the error

arr.sort(function(a, b) {
  return a - b;
});

Please read the documentation
This is from the documentation

  • If compareFunction(a, b) is less than 0, sort a to an index lower than b (i.e. a comes first).
  • If compareFunction(a, b) returns 0, leave a and b unchanged with respect to each other, but sorted with respect to all different elements. Note: the ECMAscript standard does not guarantee this behaviour, and thus not all browsers (e.g. Mozilla versions dating back to at least 2003) respect this.
  • If compareFunction(a, b) is greater than 0, sort b to an index lower than a (i.e. b comes first).
  • compareFunction(a, b) must always return the same value when given a specific pair of elements a and b as its two arguments. If inconsistent results are returned then the sort order is undefined.
1 Like
11

It works if you just comment out variables let a,b and add parentheses around a,b within the .sort((a,b) => a > b); method which is not my question.

12

the question is about parameters within the function vs separated defined variables

13

No, the error is just that you are not using parenthesis around parameters with arrow function - try changing just that and you will see, when you have more than one parameter you must use parenthesis around it

Anyway, you should use the sort method returning positive, negative, 0 - you are lucky because the browser you are using still support the true/false thing, not all do, so on different browsers you would have different results

14

That is true. but no need of vars