On Saturday 01 April 2006 11:53 am, Brian Hulley wrote:
Claus Reinke wrote:
the usual way to achieve this uses the overloading of Nums in Haskell: when you write '1' or '1+2', the meaning of those expressions depends on their types. in particular, the example above uses 'T Double', not just 'Double'.
However there is nothing in the functions themselves that restricts their use to just T Double. Thus the functions can be compared for equality by supplying an argument of type T Double but used elsewhere in the program with args of type (plain) Double eg:
Overloaded functions instantiated at different types are not (in general) the same function. If you mentally do the dictionary-translation, you'll see why. In particular for f, g :: XYZ a => a -> b, and types n m such that (XYZ n) and (XYZ m), f :: (n -> b) === g :: (n -> b) does *not* imply f :: (m -> b) === g :: (m -> b) That is where your argument falls down. Rob Dockins