Julia overwriting function before its called

user_15197182 0
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The easiest way to explain the problem is with a code snippet.

function foo()
  bar(x) = 1+x
  println(bar(1)) #expecting a 2 here
  bar(x) = -100000x
  println(bar(1)) #expecting -100000
end

foo()

OUTPUT:

-100000
-100000

I imagine that the compiler is optimizing away a function that doesn't last long, but I haven't seen anything in the docs that would cause me to expect this behavior, and Google returns nothing but the docs. What is going on here?

julia function
1 Answers

This looks like a version of https://github.com/JuliaLang/julia/issues/15602. In the upcoming Julia 1.6 release this gives a warning:

julia> function foo()
         bar(x) = 1+x
         println(bar(1)) #expecting a 2 here
         bar(x) = -100000x
         println(bar(1)) #expecting -100000
       end
WARNING: Method definition bar(Any) in module Main at REPL[1]:2 overwritten at REPL[1]:4.
foo (generic function with 1 method)

You should use anonymous functions like this instead:

julia> function foo()
         bar = x -> 1+x
         println(bar(1)) #expecting a 2 here
         bar = x -> -100000x
         println(bar(1)) #expecting -100000
       end
foo (generic function with 1 method)

julia> foo()
2
-100000
user_50871360
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