Closed mtsokol closed 3 weeks ago
This is due to how values are converted between Python and Julia.
You can do juliacall.convert(jl.Int8, 0)
to get an int
-like value which wraps the underlying Julia Int8
value. Some of this should be a bit slicker in v1 but that won't be out for a while...
Thank you for the clarification!
Affects: JuliaCall
Describe the bug
I would like to create Julia scalars by calling appropriate data-types (Int8, Int16, etc.), but instead calling specific dtypes just give python builtins (e.g.
int
for any integer julia dtype). Or is there another way to instantiate them?Your system Please provide detailed information about your system: