JavaScript reduce behavior with/without an initial value

Question

I am trying to sum the squares of numbers of an array by JavaScript reduce function. But the result differs when reduce method is called with or without the initial value.

var x = [75, 70, 73, 78, 80, 69, 71, 72, 74, 77];
console.log(x.slice().reduce((ac,n) => ac+(n*n))); // 49179
console.log(x.slice().reduce((ac,n) => ac+(n*n),0)); // 54729

This should be equivalent to the calls above:

console.log(x.slice().map(val => val*val).reduce((ac,n) => ac+n)); // 54729

In this case however both method returns the same value.

console.log([1,2,3].slice().reduce((ac,z) => ac+(z*z))); // 14
console.log([1,2,3].slice().reduce((ac,z) => ac+(z*z), 0)); // 14

Why are the results of the first two calls different and the last two the same?

Pointy · Accepted Answer · 2018-06-12 22:35:07Z

31

If you don't provide the second parameter to .reduce(), it uses the first element of the array as the accumulator and starts at the second element.

In the first example, your first result of the .reduce() iteration is

75 + 70 * 70

while in the second version where pass in an explicit 0 it's

0 + 75 * 75

The cascade of computations from that leads to different results.

In the second example, you'll end up computing

1 + 2 * 2

and then

5 + 3 * 3

in the first line, which gives 14. In the second version, when you start with 0, you'll compute

0 + 1 * 1
1 + 2 * 2
5 + 3 * 3

which is also 14.

edited Jun 12, 2018 at 22:35

answered Jun 12, 2018 at 22:32

Pointy

415k62 gold badges600 silver badges631 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

VinArrow Over a year ago

I see. So it would skip the call to (n*n). That makes sense.

Collectives™ on Stack Overflow

JavaScript reduce behavior with/without an initial value

1 Answer 1

1 Comment

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

1 Comment

Your Answer

Sign up or log in

Post as a guest

Linked

Related