1
>> f: func [x /a][either a [x + 2] [x + 1]]
== func [x /a][either a [x + 2] [x + 1]]

>> b: /a
== /a

>> f/b 1
*** Script Error: f has no refinement called b
*** Where: f
*** Stack: f  

>> f/:b 1
*** Script Error: f has no refinement called :b
*** Where: f
*** Stack: f

You can see that the function f has a refinement a, and I bind /a to b. When calling f with its refinement /a by b, it fails.

What is the correct way to pass a refinement which needs to be evaluated before to its function? Or, is there a way to convert a path! to function!?

lyl
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1 Answers1

4

Refinements are limited, in a way that they can be passed only by being listed literally in a function call. On the other hand, you can construct such function call (a path! value followed by all the arguments) whichever way you want:

>> b: 'a
== a
>> do probe reduce [to path! reduce ['f b] 1]
[f/a 1]
== 3

Note that elements of the path in this case are word!s, not refinement!s. In general, such use-case is covered by apply. However, only Rebol has it natively, and with an awkward calling convention:

>> f: func [x /a][either a [x + 2] [x + 1]]
>> b: yes
== true
>> apply :f [1 b]
== 3
>> apply :f [1 no]
== 2

You can easily write your own high-level version of it if that's what you need.

9214
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  • Perfect solution! And another method under your point: `b: "/a" do rejoin ["f" b " " 1]`. Thank you! – lyl Apr 22 '21 at 12:32