Logical expression syntactic sugar

[jw013]

Section 1.1.6 says (paraphrasing) that e1 && e2 is syntactic sugar for e1 ? e2 : false, and e1 || e2 is syntactic sugar for e1 ? true : e2. Strictly speaking, in JavaScript, e1 && e2 is syntactic sugar for e1 ? e2 : e1 and e1 || e2 is syntactic sugar for e1 ? e1 : e2.

Direct link to the relevant portion of the language specification: https://tc39.es/ecma262/multipage/ecmascript-language-expressions.html#sec-binary-logical-operators-runtime-semantics-evaluation

[martin-henz]

@jw013 sorry for the delay. I thought about this for a long time. The challenge is to not disturb the reading and learning experience by this subtle issue. I ended up referring to footnote 14 on the previous page. That allowed me to keep the wording almost intact. Instead of "This syntactic form is syntactic sugar" I use the phrase "We assume this syntactic form to be syntactic sugar", with a footnote that points out that full JavaScript needs to consider the case where the result of evaluating expression1 is neither true nor false.

By the way, your formulation is flawed in the sense that e1 ? e2 : e1 would evaluate e1 twice in case it's falsy. The actual expansion as proper syntactic sugar would need to ensure that e1 is evaluated only once, e.g. ( x => x ? e2 : x)(e1) which is of course super ugly here, as it foreshadows a technique that comes only in 1.3.2.