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JuliaLang / IJulia.jl

Julia kernel for Jupyter
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When `IJULIA_NODEFAULTKERNEL` is defined, `exe` is not. #1075

Closed jmuchovej closed 1 year ago

jmuchovej commented 1 year ago

:wave: I've run into an error while trying to build a container for a project I'm working on. We were trying to sidestep the default Julia kernel for a custom one, but it seems that exe is defined in kspec which is only loaded if IJULIA_NODEFAULTKERNEL is undefined.

Is this an intentional error?

https://github.com/JuliaLang/IJulia.jl/blob/537ee797026b8598e7a3a96e63afc4a2aaa3d27b/deps/build.jl#L3-L7

https://github.com/JuliaLang/IJulia.jl/blob/537ee797026b8598e7a3a96e63afc4a2aaa3d27b/deps/build.jl#L18

ERROR: Error building `IJulia`: 
#0 49.27 ERROR: LoadError: UndefVarError: exe not defined
#0 49.27 Stacktrace:
#0 49.27  [1] top-level scope
#0 49.27    @ ~/.julia/packages/IJulia/AQu2H/deps/build.jl:18
#0 49.27  [2] include(fname::String)
#0 49.27    @ Base.MainInclude ./client.jl:476
#0 49.27  [3] top-level scope
#0 49.27    @ none:5
#0 49.27 in expression starting at /home/vscode/.julia/packages/IJulia/AQu2H/deps/build.jl:18
stevengj commented 1 year ago

Looks like a bug, should be fixed in #1079.