350743 - Decompilation can turn a generator function into a normal (yield-less) function

Status: RESOLVED WORKSFORME

(Core :: JavaScript Engine, defect)

Description

[Jesse Ruderman]

> z = function () { print(3); if(false) yield 4; }
function () { print(3); }

Calling z() returns a generator immediately, without printing 3.  This seems consistent with http://developer.mozilla.org/en/docs/New_in_JavaScript_1.7 :

"The function containing the yield keyword is a generator. When you call it, its formal parameters are bound to actual arguments, but its body isn't actually evaluated. Instead, a generator-iterator is returned."

But the decompilation of z() does not contain "yield" at all, so calling it does print 3.  So z's decompilation is very different from z.

Comment 1

[Ryan VanderMeulen [:RyanVM]]

Current js shell results:
js> z = function () { print(3); if(false) yield 4; }
(function () {print(3);})
js> z()
({})

Comment 2

[Jesse Ruderman]

This will probably be fixed by the syntax change in bug 666399.

Comment 3

[Luke Wagner [:luke]]

It ended up being fixed by saving the source.