Passing keyword arguments through a thin wrapper without explicitly writing them out

General Usage
1

I wonder if there is a way to pass on keyword arguments through a thin wrapper without having to explicitly type the keyword arguments, something like the following Python code. Splatting doesn’t seem to work, nor does passing a dictionary, anything other obvious way I missed? Will this change with named tuples in v1.0?

>>> def f(**kwargs):
...     return kwargs
...
>>> f(a = 1, b = 2, c = 3)
{'a': 1, 'c': 3, 'b': 2}
>>> type(f(a = 1, b = 2, c = 3))
<type 'dict'>
>>> def g(**kwargs):
...     return f(**kwargs)
...
>>> g(a = 1, b = 2, c = 3)
{'a': 1, 'c': 3, 'b': 2}
2

This is what I missed :sweat_smile:

julia> f(;kwargs...) = Dict(kwargs)
f (generic function with 1 method)

julia> g(;kwargs...) = f(;kwargs...)
g (generic function with 1 method)

julia> g(a = 1, b = 2, c = 3)
Dict{Symbol,Int64} with 3 entries:
  :a => 1
  :b => 2
  :c => 3

Thanks to @mkborregaard in How to collect keyword arguments in a Dict? - #5 by mkborregaard

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3

Why do you have to recast the kwargs as a dictionary though?

It seems kind of weird you need an intermediate function that is just calling a Dict constructor?

The way I work around it is with:

default_kwargs = Dict()
f(kwargs=default_kwargs)
4

I don’t remember! It was me answering another question or some exercise or something.

5

Isn’t that just to show that the kwargs got passed through.

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