My blog does explain what reset and shift do, so you may want to read that again.
Another good source, which I also point in my blog, is the Wikipedia entry on continuation passing style. That one is, by far, the most clear on the subject, though it does not use Scala syntax, and the continuation is explicitly passed.
The paper on delimited continuations, which I link to in my blog but seems to have become broken, gives many examples of usage.
But I think the best example of the concept of delimited continuations is Scala Swarm. In it, the library stops the execution of your code at one point, and the remaining computation becomes the continuation. The library then does something -- in this case, transferring the computation to another host, and returns the result (the value of the variable which was accessed) to the computation that was stopped.
Now, you don't understand even the simple example on the Scala page, so do read my blog. In it I'm only concerned with explaining these basics, of why the result is 8.
Given the canonical example from the research paper for Scala's delimited continuations, modified slightly so the function input to shift is given the name f and thus is no longer anonymous.
def f(k: Int => Int): Int = k(k(k(7)))
reset(
shift(f) + 1 // replace from here down with `f(k)` and move to `k`
) * 2
The Scala plugin transforms this example such that the computation (within the input argument of reset) starting from each shift to the invocation of reset is replaced with the function (e.g. f) input to shift.
The replaced computation is shifted (i.e. moved) into a function k. The function f inputs the function k, where k contains the replaced computation, k inputs x: Int, and the computation in k replaces shift(f) with x.
f(k) * 2
def k(x: Int): Int = x + 1
Which has the same effect as:
k(k(k(7))) * 2
def k(x: Int): Int = x + 1
Note the type Int of the input parameter x (i.e. the type signature of k) was given by the type signature of the input parameter of f.
Another borrowed example with the conceptually equivalent abstraction, i.e. read is the function input to shift:
def read(callback: Byte => Unit): Unit = myCallback = callback
reset {
val byte = "byte"
val byte1 = shift(read) // replace from here with `read(callback)` and move to `callback`
println(byte + "1 = " + byte1)
val byte2 = shift(read) // replace from here with `read(callback)` and move to `callback`
println(byte + "2 = " + byte2)
}
I believe this would be translated to the logical equivalent of:
val byte = "byte"
read(callback)
def callback(x: Byte): Unit {
val byte1 = x
println(byte + "1 = " + byte1)
read(callback2)
def callback2(x: Byte): Unit {
val byte2 = x
println(byte + "2 = " + byte1)
}
}
I hope this elucidates the coherent common abstraction which was somewhat obfuscated by prior presentation of these two examples. For example, the canonical first example was presented in the research paper as an anonymous function, instead of my named f, thus it was not immediately clear to some readers that it was abstractly analogous to the read in the borrowed second example.
Thus delimited continuations create the illusion of an inversion-of-control from "you call me from outside of reset" to "I call you inside reset".
Note the return type of f is, but k is not, required to be the same as the return type of reset, i.e. f has the freedom to declare any return type for k as long as f returns the same type as reset. Ditto for read and capture (see also ENV below).
Delimited continuations do not implicitly invert the control of state, e.g. read and callback are not pure functions. Thus the caller can not create referentially transparent expressions and thus does not have declarative (a.k.a. transparent) control over intended imperative semantics.
We can explicitly achieve pure functions with delimited continuations.
def aread(env: ENV): Tuple2[Byte,ENV] {
def read(callback: Tuple2[Byte,ENV] => ENV): ENV = env.myCallback(callback)
shift(read)
}
def pure(val env: ENV): ENV {
reset {
val (byte1, env) = aread(env)
val env = env.println("byte1 = " + byte1)
val (byte2, env) = aread(env)
val env = env.println("byte2 = " + byte2)
}
}
I believe this would be translated to the logical equivalent of:
def read(callback: Tuple2[Byte,ENV] => ENV, env: ENV): ENV =
env.myCallback(callback)
def pure(val env: ENV): ENV {
read(callback,env)
def callback(x: Tuple2[Byte,ENV]): ENV {
val (byte1, env) = x
val env = env.println("byte1 = " + byte1)
read(callback2,env)
def callback2(x: Tuple2[Byte,ENV]): ENV {
val (byte2, env) = x
val env = env.println("byte2 = " + byte2)
}
}
}
This is getting noisy, because of the explicit environment.
Tangentially note, Scala does not have Haskell's global type inference and thus as far as I know couldn't support implicit lifting to a state monad's unit (as one possible strategy for hiding the explicit environment), because Haskell's global (Hindley-Milner) type inference depends on not supporting diamond multiple virtual inheritance.
Scala Continuations via Meaningful Examples
Let us define from0to10 that expresses the idea of iteration from 0 to 10:
def from0to10() = shift { (cont: Int => Unit) =>
for ( i <- 0 to 10 ) {
cont(i)
}
}
Now,
reset {
val x = from0to10()
print(s"$x ")
}
println()
prints:
0 1 2 3 4 5 6 7 8 9 10
In fact, we do not need x:
reset {
print(s"${from0to10()} ")
}
println()
prints the same result.
And
reset {
print(s"(${from0to10()},${from0to10()}) ")
}
println()
prints all pairs:
(0,0) (0,1) (0,2) (0,3) (0,4) (0,5) (0,6) (0,7) (0,8) (0,9) (0,10) (1,0) (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (1,7) (1,8) (1,9) (1,10) (2,0) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (2,7) (2,8) (2,9) (2,10) (3,0) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (3,7) (3,8) (3,9) (3,10) (4,0) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (4,7) (4,8) (4,9) (4,10) (5,0) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (5,7) (5,8) (5,9) (5,10) (6,0) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6) (6,7) (6,8) (6,9) (6,10) (7,0) (7,1) (7,2) (7,3) (7,4) (7,5) (7,6) (7,7) (7,8) (7,9) (7,10) (8,0) (8,1) (8,2) (8,3) (8,4) (8,5) (8,6) (8,7) (8,8) (8,9) (8,10) (9,0) (9,1) (9,2) (9,3) (9,4) (9,5) (9,6) (9,7) (9,8) (9,9) (9,10) (10,0) (10,1) (10,2) (10,3) (10,4) (10,5) (10,6) (10,7) (10,8) (10,9) (10,10)
Now, how does that work?
There is the called code, from0to10, and the calling code. In this case, it is the block that follows reset. One of the parameters passed to the called code is a return address that shows what part of the calling code has not yet been executed (**). That part of the calling code is the continuation. The called code can do with that parameter whatever it decides to: pass control to it, or ignore, or call it multiple times. Here from0to10 calls that continuation for each integer in the range 0..10.
def from0to10() = shift { (cont: Int => Unit) =>
for ( i <- 0 to 10 ) {
cont(i) // call the continuation
}
}
But where does the continuation end? This is important because the last return from the continuation returns control to the called code, from0to10. In Scala, it ends where the reset block ends (*).
Now, we see that the continuation is declared as cont: Int => Unit. Why? We invoke from0to10 as val x = from0to10(), and Int is the type of value that goes to x. Unit means that the block after reset must return no value (otherwise there will be a type error). In general, there are 4 type signatures: function input, continuation input, continuation result, function result. All four must match the invocation context.
Above, we printed pairs of values. Let us print the multiplication table. But how do we output \n after each row?
The function back lets us specify what must be done when control returns back, from the continuation to the code that called it.
def back(action: => Unit) = shift { (cont: Unit => Unit) =>
cont()
action
}
back first calls its continuation, and then performs the action.
reset {
val i = from0to10()
back { println() }
val j = from0to10
print(f"${i*j}%4d ") // printf-like formatted i*j
}
It prints:
0 0 0 0 0 0 0 0 0 0 0
0 1 2 3 4 5 6 7 8 9 10
0 2 4 6 8 10 12 14 16 18 20
0 3 6 9 12 15 18 21 24 27 30
0 4 8 12 16 20 24 28 32 36 40
0 5 10 15 20 25 30 35 40 45 50
0 6 12 18 24 30 36 42 48 54 60
0 7 14 21 28 35 42 49 56 63 70
0 8 16 24 32 40 48 56 64 72 80
0 9 18 27 36 45 54 63 72 81 90
0 10 20 30 40 50 60 70 80 90 100
Well, now it's time for some brain-twisters. There are two invocations of from0to10. What is the continuation for the first from0to10? It follows the invocation of from0to10 in the binary code, but in the source code it also includes the assignment statement val i =. It ends where the reset block ends, but the end of the reset block does not return control to the first from0to10. The end of the reset block returns control to the 2nd from0to10, that in turn eventually returns control to back, and it is back that returns control to the first invocation of from0to10. When the first (yes! 1st!) from0to10 exits, the whole reset block is exited.
Such method of returning control back is called backtracking, it is a very old technique, known at least from the times of Prolog and AI-oriented Lisp derivatives.
The names reset and shift are misnomers. These names should better have been left for the bitwise operations. reset defines continuation boundaries, and shift takes a continuation from the call stack.
Note(s)
(*) In Scala, the continuation ends where the reset block ends. Another possible approach would be to let it end where the function ends.
(**) One of the parameters of the called code is a return address that shows what part of the calling code has not yet been executed. Well, in Scala, a sequence of return addresses is used for that. How many? All of the return addresses placed on the call stack since entering the reset block.
UPD Part 2
Discarding Continuations: Filtering
def onEven(x:Int) = shift { (cont: Unit => Unit) =>
if ((x&1)==0) {
cont() // call continuation only for even numbers
}
}
reset {
back { println() }
val x = from0to10()
onEven(x)
print(s"$x ")
}
This prints:
0 2 4 6 8 10
Let us factor out two important operations: discarding the continuation (fail()) and passing control on to it (succ()):
// fail: just discard the continuation, force control to return back
def fail() = shift { (cont: Unit => Unit) => }
// succ: does nothing (well, passes control to the continuation), but has a funny signature
def succ():Unit @cpsParam[Unit,Unit] = { }
// def succ() = shift { (cont: Unit => Unit) => cont() }
Both versions of succ() (above) work. It turns out that shift has a funny signature, and although succ() does nothing, it must have that signature for type balance.
reset {
back { println() }
val x = from0to10()
if ((x&1)==0) {
succ()
} else {
fail()
}
print(s"$x ")
}
as expected, it prints
0 2 4 6 8 10
Within a function, succ() is not necessary:
def onTrue(b:Boolean) = {
if(!b) {
fail()
}
}
reset {
back { println() }
val x = from0to10()
onTrue ((x&1)==0)
print(s"$x ")
}
again, it prints
0 2 4 6 8 10
Now, let us define onOdd() via onEven():
// negation: the hard way
class ControlTransferException extends Exception {}
def onOdd(x:Int) = shift { (cont: Unit => Unit) =>
try {
reset {
onEven(x)
throw new ControlTransferException() // return is not allowed here
}
cont()
} catch {
case e: ControlTransferException =>
case t: Throwable => throw t
}
}
reset {
back { println() }
val x = from0to10()
onOdd(x)
print(s"$x ")
}
Above, if x is even, an exception is thrown and the continuation is not called; if x is odd, the exception is not thrown and the continuation is called.
The above code prints:
1 3 5 7 9
