Re: [Haskell-cafe] Generating random arguments for a function

I think it's more complicated because he doesn't know what the return type or arity of the function is. In QuickCheck they know the return type of a property is Bool. In this case, we only know that the return type is an instance of Show. I don't think that's enough to simply implement this. On Sunday, January 13, 2013, Stephen Tetley wrote:

Yes - I was just checking the first QuickCheck paper to see how the authors did this.

You would need a new type class that works like `Testable` and the versions of associated machinery `forAll` and `evaluate` to unroll function application.

On 13 January 2013 09:28, Roman Cheplyaka javascript:;> wrote:

...

This can be done with relatively simple type class hackery. In fact, QuickCheck already does that in order to generate arguments and print them in case of failure.

Roman

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