What does var = var && “*” mean in Javascript

Question

I am trying to work out some obfusicated code by reading it, and I was doing pretty well until a came across this:

a = a && \"*\"

N

Answer 1

If a is truthy, it assigns "*" to a.

If a is falsy, it remains untouched.

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&& has short-circuit semantics: A compound expression (e1) && (e2)—where e1 and e2 are arbitrary expressions themselves—evaluates to either

• e1 if e1 evaluates to false in a boolean context—e2 is not evaluated
• e2 if e1 evaluates to true in a boolean context

This does not imply that e1 or e2 and the entire expression (e1) && (e2) need evaluate to true or false!

In a boolean context, the following values evaluate to false as per the spec:

• null
• undefined
• ±0
• NaN
• false
• the empty string

All¹ other values are considered true.

The above values are succinctly called "falsy" and the others "truthy".

Applied to your example: a = a && "*"

According to the aforementioned rules of short-circuit evaluation for &&, the expression evaluates to a if a is falsy, which is then in turn assigned to a, i.e. the statement simplifies to a = a.

If a is truthy, however, the expression on the right-hand side evaluates to *, which is in turn assigned to a.

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As for your second question: (e1) || (e2) has similar semantics:

The entire expression evaluates to:

• e1 if e1 is truthy
• e2 if e1 is falsy

^{1 Exception}