How is the program executed if there are two resume?

[bitbegin]

There are two resume in effect handler, how to explain the print results?

effect<a> ctl get() : a

fun get-and-print() : <get<int>, console> ()
  println("call 1");
  val m : int = get()
  println("call 2, m: " ++ m.show());
  val n : int = get()
  println("call 3, n: " ++ n.show());
  val l : int = get()
  println("call 4, l: " ++ l.show());
  println("m: " ++ m.show() ++ " n: " ++ n.show() ++ " l: " ++ l.show() ++ "\n")

fun main()
  with ctl get()
    println("get 1");
    resume(10)
    println("get 2");
    resume(20)
    println("get 3");
  get-and-print()

output:

call 1
get 1
call 2, m: 10
get 1
call 3, n: 10
get 1
call 4, l: 10
m: 10 n: 10 l: 10

get 2
call 4, l: 20
m: 10 n: 10 l: 20

get 3
get 2
call 3, n: 20
get 1
call 4, l: 10
m: 10 n: 20 l: 10

get 2
call 4, l: 20
m: 10 n: 20 l: 20

get 3
get 3
get 2
call 2, m: 20
get 1
call 3, n: 10
get 1
call 4, l: 10
m: 20 n: 10 l: 10

get 2
call 4, l: 20
m: 20 n: 10 l: 20

get 3
get 2
call 3, n: 20
get 1
call 4, l: 10
m: 20 n: 20 l: 10

get 2
call 4, l: 20
m: 20 n: 20 l: 20

get 3
get 3
get 3

[anfelor]

Hi @bitbegin,

The resume keyword resumes the program at the point from the point where the ctl operation was called. The surprising feature of Koka here is that you can resume the program multiple times. In the first block of the output, Koka resumes the program from all get operations with the result 10. Then, in the second block, it evaluates the last get further and resumes the last get with output 20. And so forth: As you have two resumes in three get operations, the last println is executed 2^3 = 8 times. You can read more about this concept at: https://koka-lang.github.io/koka/doc/book.html#why-handlers