scala - Referring abstract type member of a type parameter -

my scenario like:

trait {     type b     def foo(b: b) }  trait c[d <: a] {     val d: d     def createb(): d#b     def bar() {         d.foo(createb)     } } 

in repl, complains

<console>:24: error: type mismatch;  found   : d#b  required: c.this.d.b        a.bar(createb()) 

what's wrong ? , (if possible @ all) how correct code ?

d#b type projection, , not same d.b. have type mismatch because in foo, bactually meant this.b, said not same d#b (the latter being more general). informally, can think of d#bas representing possible type abstract type b can take instance of d, while d.b type of b specific instance d.

see what `#` operator mean in scala? , what meant scala's path-dependent types? context.

one way make compile changing createb's return type d.b:

def createb(): d.b 

however in many cases such solution restrictive because tied specific instance d, might not had in mind. solution replace abstract type type parameter (though more verbose):

trait a[b] {   def foo(b: b) }  trait c[b, d <: a[b]] {   val d: d   def createb(): b   def bar() {     d.foo(createb)   } }