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conditional compilation
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I want to add a log function to help in debugging. But, I don't need it most of the time (and efficiency is pretty important).

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let doLogging = true

let log s = if doLogging then (* log code here *) () else ()

let myFun =

     // ...

     log (sprintf "%d" val)

     // ...

Basically, I want don't want the sprintf to be evaluated when the logging isn't happening. Is there any way to do this?

.

Brian's answer is more efficient: IIRC, the whole call to log will just disappear, including the arguments.

Another approach is to make the log function take a () -> string parameter, then call that if you want to log:

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 let loggingOn = false
let log (s:Lazy<_>) = if loggingOn then printfn "LOG: %s" (s.Force())
log (lazy sprintf "hi %s" "bye")

The downside here is that'll still construct a lazy value.

By on 8/26/2009 6:51 PM ()

If the compiler is smart, i.e. has constant propagation and dead code elimination, laziness is

unnecessary. Otherwise, even with lazy, you could end up with a function calls inside of tight loops.

By on 8/26/2009 7:42 PM ()

yanov good point. You're right, as far as I can tell, the compiler won't eliminate creation of the lazy value. However, it will elimiate the function here:

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 let loggingOn = false
let log f = if loggingOn then printfn "%s" <| f() 
log (fun () -> sprintf "hi %d" 123)

When built with loggingOn = false, the lambda isn't emitted.

By on 8/26/2009 11:43 PM ()

Yes, see

[link:msdn.microsoft.com]

For example:

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open System.Diagnostics

type MyType() =
    [<Conditional("DEBUG")>]
    static member Log s = printfn "%s" s

let F x =
    MyType.Log (sprintf "Input was %d" x)
    let r = x + 1
    MyType.Log (sprintf "Result was %d" r)
    r

printfn "%d" (F 42)

(Try running after compiling in both 'Debug' and 'Release' modes.)

By on 8/26/2009 9:11 AM ()

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