For an array (array == 0) is true and for a non-empty array (array == 0) is false why? [duplicate]

Question

let x = [];
if(x==0) 
    console.log('Hello')

The about code prints 'Hello'

let x = [ 1 ];
if(x==0) 
    console.log('Hello');

The above code does not print 'Hello'.

Why?

Many related questions that contain the answer. See Why does {} == false evaluate to false while [] == false evaluates to true? and If both [0] == 0 and 0 == [[0]] are true than why is [0] == [[0]] false?. [] == 0 and [] == false are the same because false is converted to 0. — Felix Kling
– Felix Kling, Commented Jul 14, 2017 at 23:42
[] is falsy, 0 is falsy. Falsy == Falsy === true. Pretty simple mate. — Mardoxx
– Mardoxx, Commented Jul 14, 2017 at 23:49
@Mardoxx: Except that [] is not falsey. It's not that straightforward. Have a look at the links I posted in my first comment. — Felix Kling
– Felix Kling, Commented Jul 14, 2017 at 23:51
@Mardoxx When objects of different types are compared, one of them is converted to the other type. It doesn't just compare their truthiness. — Barmar
– Barmar, Commented Jul 15, 2017 at 0:10

Felix Kling · Accepted Answer · 2017-07-15 00:07:50Z

1

While I think this should be closed as duplicate of this question, before any more false statements are made:

== performs type conversion. This is the type conversion that takes place according to the Abstract Equality Comparison Algorithm:

[] == 0                    // step 9 ToPrimitve([]) == 0
"" == 0                    // step 5 ToNumber("") == 0
0 == 0                     // step 1.c.iii

[1] == 0                   // step 9 ToPrimitve([1]) == 0
"1" == 0                   // step 5 ToNumber("1") == 0
1 == 0                     // step 1.c.iii

References: ToNumber, ToPrimitive

answered Jul 15, 2017 at 0:07

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