Scala operator #> causes compilation error but not &> - why?

Question

I have a type inference issue and asked for help here. The initial problem was due to overload. Once corrected I still had problems.

So here is the code:

class DPipe[ A ]( a: A ) {
  def !>[ B ]( f: A => B ) = Try(f( a ))
  def #>[ B, C ]( f: B => C )(implicit ev: A =:= Try[B]) : Try[C] = a.map(f)
  //def &>[ B, C ]( f: B => C )( implicit ev: A =:= Try[ B ] ) =  a.map( f )
}

object DPipe {
  def apply[ A ]( v: A ) = new DPipe( v )
}

object DPipeOps {
  implicit def toDPipe[ A ]( a: A ): DPipe[ A ] = DPipe( a )
}

And here are the tests:

object DPipeDebug {

 def main( args: Array[ String ] ) {

    import DPipeOps._

    val r8 = 100.0 !> {x : Double => x / 0.0}  
    println(r8)
    val r9 = r8 #> {x:Double => x* 3.0} 
    println(r9)
    /*
    val r8 = 100.0 !> { x: Double => x / 0.0 }
    println( r8.get )
    val r9 = r8 &> { x: Double => x * 3.0 }
    println( r9 )*/

    val r10 = (100.0 !> {x : Double => x / 0.0}) #> {x:Double => x* 3.0} 
   //val r10 = ( 100.0 !> { x: Double => x / 0.0 } ) &> { x: Double => x * 3.0 }

    val r11 = 100.0 !> {x : Double => x / 0.0} #> {x:Double => x* 3.0} 
    //val r11 = 100.0 !> { x: Double => x / 0.0 } &> { x: Double => x * 3.0     }
  }

}

As it stands we have the following error in the last code line:

Cannot prove that Double => Double =:= scala.util.Try[Double].
val r11 = 100.0 !> {x : Double => x / 0.0} #> {x:Double => x* 3.0} 
                                           ^

Notice that in the second last code line, I need only add the parenthesis to enforce left left-hand associativity (Scala default). It seems like the #> operator tries to use the function {x : Double => x / 0.0}, which indeed is a Double.

If however I use the "&>" operator, no error occurs. In the test code below, just flip the comments. So my question is, why is this happening. Is this something new to Scala 2.12.0?

TIA

@Łukasz - I did not consider operator precedence. Blinded by the overload issue which led me to consider operator associativity. So this is the explanation I needed: # has higher precedence than ! (as per the link you made available). So can you make that an answer so that I can tag it as the correct one? TIA. — user2051561

Łukasz Łukasz · Accepted Answer · 2016-11-24T17:22:25

The problem is with the operator precedence. Here are more details: Operator precedence in Scala

It depends on the first character of the operator. # has higher precedence than ! and ! has higher precedence than &.

Hence you get

100.0 !> ({x : Double => x / 0.0} #> {x:Double => x* 3.0})

instead of

(100.0 !> {x : Double => x / 0.0}) #> {x:Double => x* 3.0}

Scala operator #> causes compilation error but not &> - why?

1 Answers