Why does !!1==“1” equal true and !!2==“2” equal false?

Question

As the title states, why does:

> !!1==\"1\"

equal

True

and

> !!2==\"2\"
         


        

Answer 1

tldr; this is due to the [ToNumber] conversions in the == operator algorithm.

The first step is to simplify the expression. Since !!x=="x" is parsed like (!!x)=="x" and !!a_truthy_expression -> true, the actual relevant expression for the equality is

!!1=="2" -> true=="1" -> Boolean==String
!!2=="2" -> true=="2" -> Boolean==String

So then looking at the rules for 11.9.3 The Abstract Equality Comparison Algorithm and following along with the application yields

Rule 6 - If Type(x) is Boolean, return the result of the comparison ToNumber(x) == y.

which results in Number==String or 1=="1" and 1=="2", respectively¹. Then the rule

Rule 7 - If Type(x) is Number and Type(y) is String, return the result of the comparison x == ToNumber(y).

is applied which results in Number==Number or 1==1 and 1==2, respectively¹; the latter is clearly false.

Rule 1 - If Type(x) is the same as Type(y), then [by c.iii.] If x is the same Number value as y, return true [else return false].

(The same algorithm explains the String==Boolean case when the complementing rules are applied.)

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¹To see the [ToNumber] rule applied, consider:

+false -> 0
+true  -> 1
+"1"   -> 1
+"2"   -> 2