Wrong slurp flag for slurps with default args

[untoreh]

## Passes
args = splitdef(Meta.parse("f(v::Int...) = v"))[:args]
spa = splitarg(args[1])
@assert spa[3] == true 
@assert spa[4] == nothing

## Slurp fails
args = splitdef(Meta.parse("f(v::Int...=0) = v"))[:args]
spa = splitarg(args[1])
@assert spa[3] == true # This one fails, should be true
@assert spa[4] == 0

[cstjean]

I don't get it. What behaviour did you expect?

[untoreh]

the third assertion should pass, v is a slurp argument with a default value, so the third element of the tuple returned by splitarg should return true since it is the slurp flag (name, type, SLURP, default)

[cstjean]

Oh, ok, thanks for the edit. I agree, something's wrong here. PR welcome.