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match inside a match?
f#

Hi, group

I have a question about using matching. If i have two input variables A and B. A can be 1, 2. B can be 1 to 20.

Here is what I want to get:

if (A = 1 && B between 1 to 10) then print "good"

if (A = 2 && B between 11 to 20) then print "bad"

how can i use match to solve this? I do not want to list all the combinations.

Thanks.

.

Here you can use the following code:

match A with

| 1 when B < 10 && B > 1 -> "good"

| 2 when B > 11 && B < 20 -> "bad"

By on 2/12/2009 10:15 PM ()

Hi, you can also use active patterns and write something like this:

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// pattern to check whether the number 'x' is in the specified range
let (|Between|_|) (nfrom, nto) x = 
  if x < nfrom && x > nto then Some(x) else None

match A, B with
| 1, Between(1, 10) _ -> "good"    // note: the underscore ignores the value returned from the pattern
| 2, Between(11, 20) _ -> "good"   // (which is the original number in this example)

I believe this makes the code a bit more readable :-).
T.

By on 2/13/2009 3:29 AM ()

You can rid of the underscores by returning Some () instead of Some(x) from the active patternThen it's even more readable [;)]

By on 2/13/2009 4:13 AM ()

That doesn't seem to work unfortunatelly - the active pattern returns a unit value that you still have to consume somehow, so if I write Some() I still have to write something, either underscore or this:

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match 0, 0 with
| 1, Between(1, 10) () -> "good"
| 2, Between(11, 20) () -> "good"

Perhaps active patterns should be able to return just true/false in this case...

T.

By on 2/13/2009 4:25 AM ()

Hmm, it doesn't work indeed. The problem apparently isn't the return value, but the fact that active patterns do not behave like curried functions. This for instance works fine:

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let (|Between_1_and_10|_|) x =
  if x>=1 && x<=10 then Some() else None
  
match 0 with
| Between_1_and_10 -> "good"
| _ -> "bad"
By on 2/13/2009 5:27 AM ()

Unless you are going out of your way to be recondite, the obvious way to write this is to use a simple if..then:

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let a = 1
let b = 10
if (a = 1 && 1<=b && b<=10) then printfn "good"
if (a = 2 && 11<=b && b<=20) then printfn "bad"
By on 2/13/2009 5:21 AM ()

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