Question

Can anyone explain why here a = [] ? 1 : 2 a will be equal to 1 and here b = [] == true ? 1 : 2 b will be equal to 2

Answer 1

A very similar case is handled in Why does ('0' ? 'a' : 'b') behave different than ('0' == true ? 'a' : 'b').

In short:

• When you have a single value that is evaluated in a boolean context, it is passed to ToBoolean to be converted to a boolean value.

• If you compare values with == though, the algorithm in section 11.9.3 of the ECMAScript specification takes place. In this algorithm, both operands are converted to either strings or numbers and these values are compared.

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More detailed explanation

a = [] ? 1 : 2

The condition is converted to a boolean by calling ToBoolean(GetValue(lref)). Now I assume that for objects, GetValue just returns the object itself and all objects evaluate to true.

Since arrays are just objects, the result of this expression is 1.

b = [] == true ? 1 : 2

As already said, quite some type conversion is going on here.

First, true is converted to a number:

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

false is converted to 0 and true is converted to 1.

So we have

[] == 1

Then step 9 applies:

9. If Type(x) is Object and Type(y) is either String or Number,
return the result of the comparison ToPrimitive(x) == y.

That means [] is converted to a primitive value (string or number). If you follow the ToPrimitive function (and section 8.12.8), you’ll find out that arrays are converted to strings, by concatenating their elements with ,. Since an empty array as no elements, the result is an empty string.

So we end up with

"" == 1

Then step 5 applies:

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

An empty string is converted to 0, hence 0 == 1 is false and the conditional operator returns 2.