CODEWITHSUNDEEP
javascriptecmascript-6
Tychio
Tychio

Reputation: 627

Why Math.max(...[]) is equal to -Infinity in ES2015?

Math.max([]) would be 0

And [..[]] is []

But why Math.max(...[]) is equal to -Infinity in ES2015?

Upvotes: 18

Views: 11146

Answers (6)

Coach Tim
Coach Tim

Reputation: 11

When finding out what the biggest number is, where do you start? The answer is the lowest number, -Infinity

function max (array) {
    var highest = -Infinity;

    foreach(elem in array) if (elem > highest) highest= elem;

    return highest;
}

Not accurate code but you get the gist, credit to my friend since the internet let me down once again.

Upvotes: 1

Struggler
Struggler

Reputation: 682

FTR the way to get around this is to use a MIN value with the spread operator. Like:

Math.max(MIN_VALUE, ...arr)
Math.max(0, ...[]) --> 0
Math.max(0, ...[1]) --> 1
Math.max(0, ...[23,1]) --> 23

Upvotes: 4

Me.Name
Me.Name

Reputation: 12544

What happens with Math.max([]) is that [] is first converted to a string and then to a number. It is not actually considered an array of arguments.

With Math.max(...[]) the array is considered a collection of arguments through the spread operator. Since the array is empty, this is the same as calling without arguments. Which according to the docs produces -Infinity

If no arguments are given, the result is -Infinity.

Some examples to show the difference in calls with arrays:

console.log(+[]); //0    [] -> '' -> 0
console.log(+[3]); //3    [] -> '3' -> 3
console.log(+[3,4]); //Nan 
console.log(...[3]); //3
console.log(...[3,4]); //3 4 (the array is used as arguments)
console.log(Math.max([])); //0  [] is converted to 0
console.log(Math.max()); // -infinity:  default without arguments
console.log(Math.max(...[])); // -infinity
console.log(Math.max([3,4])); //Nan
console.log(Math.max(...[3,4])); //4

Upvotes: 12

Rajaprabhu Aravindasamy
Rajaprabhu Aravindasamy

Reputation: 67207

If you see the internal implementation documentation, you can tell why Math.max is returning -Infinity when there is no argument passed.

If no arguments are given, the result is -∞.

So when you spread an empty array in a function call, it is like calling the function without an argument.

Upvotes: 4

towc
towc

Reputation: 2034

If you look at the babel output for Math.max(...[]), you end up with Math.max.apply(Math, []). If you log that in ES5, you see that for some reason it gives you -Infinity, because it's the same as calling it without an argument.

And indeed, Math.max() gives -Infinity

If you need a reminder: fn.apply( yourThis, [ a, b, c ] ) is the same as fn( a, b, c )

Upvotes: 3

Roman Hocke
Roman Hocke

Reputation: 4239

Because Math.max(...[]) is not Math.max([...[]]). In the first case, what You really call is Math.max(), which is -Infinity. See the spread operator in function call - https://developer.mozilla.org/cs/docs/Web/JavaScript/Reference/Operators/Spread_operator

Upvotes: 2

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