Running this.
(define f (case-lambda*
((x #:optional y) 1)
((x #:key y) 2)
((x y #:key z) 3)))
(f 1 2)
(f #:y 2)
(f 1 2 #:z 3)
I get.
Backtrace:
In ice-9/boot-9.scm:
157: 7 [catch #t #<catch-closure 8f15df0> ...]
In unknown file:
?: 6 [apply-smob/1 #<catch-closure 8f15df0>]
In ice-9/boot-9.scm:
63: 5 [call-with-prompt prompt0 ...]
In ice-9/eval.scm:
414: 4 [eval # #]
In ice-9/boot-9.scm:
2131: 3 [save-module-excursion #<procedure 8ed31c0 at
ice-9/boot-9.scm:3660:3 ()>]
3667: 2 [#<procedure 8ed31c0 at ice-9/boot-9.scm:3660:3 ()>]
In unknown file:
?: 1 [load-compiled/vm
"/home/german/.cache/guile/ccache/2.0-LE-4-2.0/home/german/Escritorio/test.scm.go"]
In /home/german/Escritorio/./test.scm:
3: 0 [f 1 #<undefined> 2 #:z 3]
/home/german/Escritorio/./test.scm:3:10: In procedure f:
/home/german/Escritorio/./test.scm:3:10: In procedure #<procedure f (x
#:optional y) | (x #:key y) | (x y #:key z)>: Odd length of keyword
argument list
But there is one interest line
3: 0 [f 1 #<undefined> 2 #:z 3]
What mean this? That is taking the second argument as a keyword?
