I got question today, that I couldn't answer. I would appreciate if you could just explain it to me. Why my array doesn't output [1, 2, 3, 4, 5] but only [2, 3, 4, 5]? This is the code:
let numbers = [1, 2, 3, 4, 5]
let order = numbers.sort((a, b) => {
console.log(a);
});
But when I use this code, it works perfectly fine
let numbers = [1, 2, 3, 4, 5]
let order = numbers.sort((a, b) => {
return a;
})
console.log(order);
It doesn't output [2, 3, 4, 5] it outputs 2, 3, 4 and 5 each in a separate step of the iteration. If you also log order you'll see that the array still contains [1, 2, 3, 4, 5]. The comparator function receives two elements of the array that it compares to each other. That way it knows in which order those elements should be. If you only log a and not b, you'll wont see all the information that is passed to the comparator. If you log both a and b, you'll see that also 1 is passed, to parameter b.
let numbers = [1, 2, 3, 4, 5]
let order = numbers.sort((a, b) => {
console.log('Comparing', a, b);
});
console.log('Final result', order);Keep in mind that the comparator function should return either a positive number when a is greater than b, a negative number when a is less than b and 0 when both elements are equal to each other. The above comparator function doesn't return any value at all, so it might not do what you expect it to do.
(Also your snippet that "works perfectly fine" doesn't follow that contract. It returns a positive number for each element, so it is basically saying that every element is greater than every other element.)
The sort() method sorts the elements of an array in place and returns the sorted array.
[1, 2, 3, 4, 5].sort((a, b) => {
console.log({a, b}); // a is secondItem, b is firstItem of an array
return -1; // return -ve value for DESC and truthy value for ASC order
});
Output will be
{a: 2, b: 1}
{a: 3, b: 2}
{a: 4, b: 3}
{a: 5, b: 4}
sort output: [5, 4, 3, 2, 1]
