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fix: PhpCall (PhpVar) translating to 'name' instead of '$name'

Open entropitor opened this issue 2 years ago • 1 comments

I was getting outputs where the code was calling

$fn($arg1, $arg2)

instead of

fn($arg1, $arg2)

(which caused a problem because the variable $fn doesn't exist)

Probably because this line is commented out https://github.com/thinkbeforecoding/peeble/blob/244c6f5ed87b9d96755b2fd44ea710c0dbcbd3db/src/peeble/Transforms.fs#L404

entropitor avatar Jan 07 '22 20:01 entropitor