Scala named and default arguments in conjunction with implicit parameters

Question

Consider the following:

def f(implicit a: String, y: Int = 0) = a + ": " + y
implicit val s = "size"
println(f(y = 2))

The last expression causes the following error:

not enough arguments for method f: (implicit a: String, implicit y:
Int)java.lang.String. Unspecified value parameter a.

However, if you provide a default value to the implicit parameter a, there is no issue:

def f(implicit a: String = "haha!", y: Int = 0) = a + ": " + y
implicit val s = "size"
println(f(y = 2))

But the last line prints

haha!: 2

while I would have expected

size: 2

So the implicit value 's' is not picked up. If you instead don't provide any parameters to f and just call

println(f)

then the implicit value is picked up and you get

size: 0

Can someone shed some light on what's going on here?

jsuereth · Accepted Answer · 2012-02-23 16:38:45Z

16

Try

println(f(y = 2, a = implicitly))

Once you start specifying parameters, you can't go back. It's either the whole list is implicit or none of it is.

answered Feb 23, 2012 at 16:38

jsuereth

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Comments

Vadim Samokhin · Accepted Answer · 2013-01-07 11:41:44Z

4

Implicit parameters should go separately -- first, and in the end of method definition -- second. Like this:

def f(y: Int = 0)(implicit a: String) = a + ": " + y
implicit val s = "size"
println(f(y = 2))

Ouputs

size: 2

answered Jan 7, 2013 at 11:41

Vadim Samokhin

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1 Comment

raisercostin Over a year ago

To use first default parameter must call f() .

Dylan · Accepted Answer · 2012-02-23 22:02:14Z

0

Along the lines of what jsuereth said, you could define your function as

def f(a: String = implicitly, y:Int = 0) = a + ": " + y

Or in the way I'm more used to seeing,

def f(y:Int = 0)(implicit a: String) = a + ": " + y

answered Feb 23, 2012 at 22:02

Dylan

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1 Comment

jsuereth Over a year ago

You should check which implicit scope that implicitly is using. I don't think it's the same as the second option.

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