I came across this strange behavior: function f(k) if k == 1 g(k) = "k = 1" elseif k == 2 g(k) = "k = 2" end g(k) end f(1) # returns "k = 2" f(2) # errors ERROR: UndefVarError: g not defined What’s going on here? How can I get around this?

[image] Confused by behaviour of nested functions New to Julia Don’t define functions conditionally inside other functions: disallow methods of local functions in different blocks · Issue #15602 · JuliaLang/julia · GitHub . and messages below

Wow, good find! Seems it’s a pretty old issue.

Function inside function is not defined?

General Usage

giordano March 2, 2021, 2:53am 2

and messages below

Topic		Replies	Views
Strange behavior when defining a function inside an `if` General Usage	2	346	August 28, 2020
Confused by behaviour of nested functions New to Julia	6	2359	February 8, 2019
Function that's `not defined` is defined General Usage question	3	676	November 16, 2020
Return a function defined in a function New to Julia	3	898	March 14, 2019
Same code in if and else blocks, get UndefVarError only on else condition General Usage scope	3	917	June 13, 2018

Function inside function is not defined?

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