[Haskell-beginners] truncate results depend on strict/lazy

[Mihai Maruseac]

On Mon, Sep 9, 2013 at 1:26 PM, Bryan Vicknair <bryanvick at gmail.com> wrote:
> Deep in a WAI web app, I have a function that converts a String from a web form
> like "0.12" to an Int 12.  Converting "0.12" to the Float 0.12 is working.
> However, converting the Float 0.12 to the Int 12 does not work as expected
> unless I use the trace function.

Not answering your question but can't you read . drop 1 . dropWhile
(/= '.') $ "0.12" ?

>
> In the following, f = 0.12::Float, gotten from a function that parses "0.12"
> into 0.12.
>
> In the following expression, the result is: Success (Just 11).
>
>> Success $ Just $ truncate (f * 100)
>
> In the following expression, the result is: Success (Just 12)
>
>> let expanded =  f * 100
>>     ans      = truncate expanded
>> in trace (show expanded) $ Success $ Just $ ans
>
> That made me think that "f * 100" had to be strictly evaluated before given to
> truncate for some reason, so I tried using seq to get the same effect, but that
> didn't work.  Am I correct in assuming that laziness has something to do with
> this problem?
>
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-- 
MM
"All we have to decide is what we do with the time that is given to us"