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fscheck generating string with size between min & max

I try to write a FsCheck generator that generates strings with length in a given interval.

My attempt is the following:

let genString minLength maxLength = 
    let isValidLength (s : string) = 
        s.Length >= minLength && s.Length <= maxLength

    Arb.generate
    |> Gen.suchThat isValidLength
    |> Arb.fromGen

...and I get the error:

"System.Exception : No instances of class FsCheck.Arbitrary`1[a] for type System.String with arguments set []"

What am I doing wrong?

thanks!

UPDATE 1:

I managed to write the generator like this:

let genStrings minLength maxLength = 
    gen {
        let! length = Gen.choose (minLength, maxLength)
        let! chars = Gen.arrayOfLength length Arb.generate<char>
        return new String(chars)
    }

Is there a better way?

UPDATE 2: I wanted to add this as a separate question but it's pretty much the same issue as my original one.

So I refactored the above code to the following structure in order to reuse the sequence generator:

let seqOfLength lengthInterval generator =
    gen {
        let! length = Gen.choose lengthInterval
        let! items = Gen.arrayOfLength length generator
        return items |> Seq.ofArray
    }

let sizedString lengthInterval =
    seqOfLength lengthInterval Arb.generate<char>
    |> Gen.map Strings.ofCharSeq

Now I'm getting the runtime error:

System.Exception : No instances of class FsCheck.Arbitrary`1[a] for type System.Char with arguments set []

...which brings me back to my original issue: Why do can't it find any instance of Arbitrary for System.Char? I thought the arbitrary for the basic types are registered by default. What am I doing wrong?

Thanks!

like image 971
vidi Avatar asked Oct 08 '14 11:10

vidi


Video Answer


1 Answers

Your sample code works as of FsCheck v2.14.3

let seqOfLength lengthInterval generator =
    gen {
        let! length = Gen.choose lengthInterval
        let! items = Gen.arrayOfLength length generator
        return items |> Seq.ofArray
    }
let stringOfLength lengthInterval =
    seqOfLength lengthInterval Arb.generate<char>
    |> Gen.map String.Concat
like image 186
DharmaTurtle Avatar answered Oct 16 '22 15:10

DharmaTurtle