Are Vararg only allowed as a final argument and if so why?

General Usage
1

For example

julia> f(x::Vararg{Int,N}, y::Float64) where {N} = (x, y)

ERROR: ArgumentError: Vararg on non-final argument in method definition for f at REPL[9]:1

In principle the method signature is not ambiguous so it’s not clear why this doesn’t work.
Is this just a temporary limitation or more fundamental ?

2

Quoth the Jeff:

This is absolutely not supported yet. It would be very useful, but extremely difficult.

2 Likes
3

It does not work because Vararg is defined to work if and only if it is the last parameter.
To pass some varying number of values as your x, use a Tuple (1, 2) (1, 2, 3) etc
where f(x::Tuple, y::Float64).

4

It isn’t really a limitation since you can’t have more than one vararg in the method signature, no matter what. Having it in a location other than the end will only add unnecessary complexity.

5

Actually there are use cases, see for example https://github.com/JuliaLang/julia/pull/31654

the API would be more uniform in that case if the size argument came last, but it seems it is not possible since it would require a vararg as a first argument.

In any case, it should at least be easiest to allow to be either the first or last argument in a method.

6

The issue is closed, but the problem is not. Maybe accidental?

7

The issue was closed by fixing the other bug it listed and improving the error message.