Make function arguments positional or keyword

General Usage
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21

@mkborregaard: I just wanted to have that flexibility in case people would ask. I love Julia language, and I would like to give people no chance to discard the fact that this is an awesome language Thanks.

22

Itā€™s not about the REPL or Atom. It is simply that there are two ways of defining a function:

f(x) = println(x)
# or
function f(x)
    println(x)
end

They are equivalent. But you tried a mix between the two, e.g. you tried something like this:

f(x)
   println(x)
end

That doesnā€™t work.

2 Likes
23

Iā€™ve actually written a little package called RequiredKeywords.jl that implements this solution using a macro. You can use it like

julia> using RequiredKeywords

julia> @required_keywords f(x; y) = x*y
f (generic function with 1 method)

julia> f(2,y=2)
4

julia> f(2)
ERROR: Unassigned Keyword:  Required keyword y::Int64 not included.
Stacktrace:
 [1] f(::Int64) at ./REPL[5]:1

It also works for the longform function style definitions, and you can still specify types for the optional keyword arguments.

1 Like
24

@adamslc: Great. Thanks.

25

Just a quick question. When we write function this way, how do we handle function if the code lines extend multiple lines, but not s single one?
Thanks.

26 
f(x) = (y = x + x; println(y))
27

@mkborregaard: OK. Thanks.

28

Or with beginā€¦ end

2 Likes
29

I think I saw somewhere that

f(x) = begin
    y = x + x
    println(y)
end

was deprecated as recommended style in favour of

function f(x)
    y = x + x
    println(y)
end

so I thought only the really compact format was recommended?

30

I donā€™t think the former was ever recommended style, but I donā€™t see anything wrong with it either. It can be nice for consistency if a file primarily contains pure functions defined from short form syntax. Scala encourages the = syntax for pure functions too, and I think thatā€™s a good convention: it is weird to see foo!(x) = bar!(x) because mutation is not typically mathematically expressed in definitions.

1 Like
31

How could I reduce the length of my function definitions if I have multiple named arguments? For example, I want to do something like:

err = z -> throw(ArgumentError("missing x"))
f(;x::Int64=err) = x
f()

but this gives a method-not-found error.

I see that there are easier ways for v1 (i.e., just doing f(; x::Int)), but how could I do for v0.6?

EDIT: can do with

err() = throw(ArgumentError("missing x"))
f(;x::Int64=err()) = x
f()
32

Just use the Compat package and put @compat in front of the function declaration, and then you can use the Julia 1.0 syntax for required-keyword arguments.

3 Likes
33

3 posts were split to a new topic: Struct with fixed array sizes

34

A post was merged into an existing topic: Struct with fixed array sizes

35

Sorry to revive this. Using Compat 1.3.0, and Julia v0.6.4, I get the following error:

julia> using Compat
INFO: Recompiling stale cache file /homes/USER/.julia/lib/v0.6/Compat.ji for module Compat.

julia> Pkg.status("Compat")
 - Compat                        1.3.0

julia> @compat f(; x::Int64) = x
ERROR: syntax: invalid keyword argument syntax "x::Int64" (expected assignment)

julia> @compat function(; x::Int64)
           return x
       end
ERROR: syntax: invalid keyword argument syntax "x::Int64" (expected assignment)

What am I messing up?

36

Looks like a https://github.com/JuliaLang/Compat.jl/pull/586 ā€” it didnā€™t handle the case of typed keyword arguments.

1 Like