After reading Chris' answer to F# - public literal and the blog post at http://blogs.msdn.com/b/chrsmith/archive/2008/10/03/f-zen-the-literal-attribute.aspx I don't get why the following is not working:
[<Literal>]
let one = 1
[<Literal>]
let two = 2
let trymatch x =
match x with
| one -> printfn "%A" one
| two -> printfn "%A" two
| _ -> printfn "none"
trymatch 3
This keeps printing "3", although I think it shouldn't. What is it that I don't s开发者_开发知识库ee here?
I think that literals need to be Uppercase. The following works fine:
[<Literal>]
let One = 1
[<Literal>]
let Two = 2
let trymatch x =
match x with
| One -> printfn "%A" One
| Two -> printfn "%A" Two
| _ -> printfn "none"
trymatch 3
In addition, if you want a nice general solution for this without using literals, you can define a parameterized active pattern like this:
let (|Equals|_|) expected actual =
if actual = expected then Some() else None
And then just write
let one = 1
let two = 2
let trymatch x =
match x with
| Equals one -> printfn "%A" one
| Equals two -> printfn "%A" two
| _ -> printfn "none"
The other answers are right - you must start your identifier with an uppercase letter. See section 7.1.2 of the spec (Named Patterns), which states that:
If long-ident is a single identifier that does not begin with an uppercase character then it is always interpreted as a variable-binding pattern and represents a variable that is bound by the pattern
Also if you don't want to have Uppercase literals you can put them in a module (here named Const):
module Const =
[<Literal>]
let one = 1
[<Literal>]
let two = 2
let trymatch x =
match x with
| Const.one -> printfn "%A" Const.one
| Const.two -> printfn "%A" Const.two
| _ -> printfn "none"
trymatch 3
Don't ask me why, but it works when you write your literals uppercase:
[<Literal>]
let One = 1
[<Literal>]
let Two = 2
let trymatch (x:int) =
match x with
| One -> printfn "%A" One
| Two -> printfn "%A" Two
| _ -> printfn "none"
trymatch 3
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