On Wed, 28 Dec 2011 22:23:12 -0600 Antoine Latter wrote:

On Wed, Dec 28, 2011 at 8:54 PM, Mike Meyer wrote:

...

While haskell's type system is usually a delight to work with. However, every once and a while I need to do mixed mode programming in spite of Kernighan and Plauger's advice, and wind up cursing the numeric type stack. I was wondering it there was a writeup on it somewhere? Preferably one aimed at practical programming. A chapter in Real World Haskell would have been ideal, but it doesn't seem to exist. For shear comprehensiveness, there's always the language report: http://www.haskell.org/onlinereport/haskell2010/haskellch6.html#x13-1350006....

Not to bad. Then again, I'm a long-time language lawyer, so others may disagree.

I don't know of anything that lays out and introduction with best practices and common pitfalls.

Given reading the report, and a little experimentation, I figured out the problem I was having. I probably will in the future as well. I wanted a percentage value rounded to the nearest integer. No problem, it's just: (%) a b = round $ 100 * a / b And that works in ghci. However, it has the type (%) :: (RealFrac a, Integral b) => a -> a -> b Which means that the argument from length was of the wrong type for this. And naturally, declaring the argument types that matched the result from length caused type problems inside (%). So, is there a reasonable way to get the value of two Integral types divided by each other and rounded? How about one integral type and one RealFrac? I know I can get it truncated towards either 0 or negative infinity, but that's not what I want here. Thanks, http://www.mired.org/ Independent Software developer/SCM consultant, email for more information. O< ascii ribbon campaign - stop html mail - www.asciiribbon.org

On Thu, 29 Dec 2011 04:49:53 -0500 Brandon Allbery wrote:

On Thu, Dec 29, 2011 at 00:33, Mike Meyer wrote:

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So, is there a reasonable way to get the value of two Integral types divided by each other and rounded? How about one integral type and one RealFrac? I know I can get it truncated towards either 0 or negative infinity, but that's not what I want here. You have two options: (1) use fromIntegral to coerce an Integral value to a RealFrac;

Yup. that's what I did.

(2) use (div) (division on Integrals) instead of (/).

Not an option, because div doesn't round, it truncates. Figuring out whether to add one to get it to round would be a lot messier than just using fromIntegral. The other half of this is that one of the two values is computed from constants in the code. So the code works if I don't coerce it. But all the computation for that value is done using floats. Seems like a violation of POLA to me. http://www.mired.org/ Independent Software developer/SCM consultant, email for more information. O< ascii ribbon campaign - stop html mail - www.asciiribbon.org