What `if ([ ])` turns out to be,it seems not falsy

Question

I got an empty array from previously code running,and I wanted to check whether it was empty within an if statement,I used

if (emptyArr) {
  //do something here. 
}

at first,received some weird results,now I know that I should use if (emptyArr.length === 0 ) {} instead,but only for curiosity,what the code block above results in?

Have you tried running `var foo = []; if (foo) { console.out("bar"); } else { console.out("baz"); }`? — fmt, Sep 08 '16 at 12:56
`!![]` → `true` – In other words, an empty array is still *truthy*. — deceze, Sep 08 '16 at 12:56
If it is different than `0`, `undefined`, `false`, `null` (i.e. falsy values), then it is the same as true. — Manuel Reis, Sep 08 '16 at 12:56

score 2 · Accepted Answer · answered Sep 08 '16 at 12:56

2

An empty array is still an object, and an object reference is always true in an if predicate like that. Thus

if ([]) alert("hello world");

will always trigger the alert.

answered Sep 08 '16 at 12:56

Pointy

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Solution: use `length`. – Tushar Sep 08 '16 at 12:56
1

@Tushar yes, as the OP says. – Pointy Sep 08 '16 at 12:57

score -1 · Answer 2 · edited May 23 '17 at 12:08

-1

typeof []
"object"

So as if({}) turns out to be true, if([]) is true too.

Here is question about Why object is true

edited May 23 '17 at 12:08

Community

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answered Sep 08 '16 at 12:57

Ivan Shmidt

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What `if ([ ])` turns out to be,it seems not falsy

2 Answers2