On 2015/06/30 18:11:05, binji wrote:
On 2015/06/30 at 17:42:43, jarin wrote:
> I do not think this is doing what you think it's doing. E.g.
>
> var x = false;
> if (x) function f() { print("f"); }
> f();
>
> still happily prints f.
>
> You really have to say
> var f = function () { ... }
>
> Even then I am not entirely sure why this has to be conditioned on
this.Worker
being defined?
Oh, I wasn't trying to prevent the definition of f, it's just that
sometimes
Worker isn't defined, so I was wrapping the whole test in the Worker
check.
I'll
move f out to make that clearer.
You are right, I indeed somehow misread the code. Please, ignore this
comment.
https://codereview.chromium.org/1216023003/
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