Open mmikhasenko opened 1 month ago
Ah, that's slightly awkward. On one hand, we could do striplines
here (and/or remove the begin block) and it would work fine. On the other hand, expr
is technically a block that contains a function definition; not a function definition.
I would lean towards lining up the behaviour with splitdef
. Does splitdef(expr)
work?
yes, splitdef
works. It correctly identifies the body and args
Then let's fix isdef
! Thank you for the bug report. I think it's basically just to remove the block. There's already a function for it in utils.jl IIRC
I need to convert a mathematical function in a string into Julia lambda function. Just found the package, it seems very useful.
Just stumbled by a very basic example:
expr = :(x -> x + 2)
is recognized as a function byisdef
, whilequote x -> x + 2 end
is not recognized.MWE
julia v1.10 MacroTools v0.5.13