Predict the Result: Type Coercion

What happens when you add "0" to a number? Evaluate each expression:

1 + "0"            1 - "0"            1 + +"0"           "5" * "2"
1 + null           1 + undefined      true + true        [] + {}
[1, 2, 3] + [4]    0 == ""            "0" == ""          null == 0
null == undefined  [] == ![]          "10" > "9"         10 > "9"
0.1 + 0.2 === 0.3  NaN === NaN        typeof null

Hints

Hint 1

+ concatenates as soon as either side becomes a string; every other arithmetic operator converts to numbers.

Hint 2

Under ==, null equals only undefined, and objects are converted to primitives before comparing.

Solution

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Answer

The rule behind all of them: + concatenates if either side becomes a string, and every other arithmetic operator converts to numbers.

Expression	Result	Expression	Result
`1 + "0"`	`"10"`	`"0" == ""`	`false`
`1 - "0"`	`1`	`null == 0`	`false`
`1 + +"0"`	`1`	`null == undefined`	`true`
`"5" * "2"`	`10`	`[] == ![]`	`true`
`1 + null`	`1`	`"10" > "9"`	`false`
`1 + undefined`	`NaN`	`10 > "9"`	`true`
`true + true`	`2`	`0.1 + 0.2 === 0.3`	`false`
`[] + {}`	`"[object Object]"`	`NaN === NaN`	`false`
`[1, 2, 3] + [4]`	`"1,2,34"`	`typeof null`	`"object"`
`0 == ""`	`true`

Two strings compare character by character, so "10" > "9" is false. null equals only undefined under ==. [] == ![] is true because ![] is false, and both sides then become 0.

Asked at

Apple