On Tue, Apr 10, 2012 at 20:53, <j.romildo@gmail.com> wrote:
On Tue, Apr 10, 2012 at 02:16:15PM -0700, Tim Perry wrote:
> I believe that "f 0 = ..." is a guard and the guard is pattern matching on
> the constructor. Despite the fact that you don't have an instance of "f _ =
> ....", the compiler needs an Eq instance to determine if it should run the
> "f 0" version of the function.
>
> Does that make sense? Hopefully someone with a better grasp of the topic
> will fill in the details.

I think you are using the wrong terms. The given examples does not make
any use of guards. Guards are boolean expressions attached to the right

It is using guards; you don't see them, because it's quietly translated by the compiler in the same way that `do` blocks are translated into applications of (>>=) and (>>) operators.

The reason for this is that numeric literals aren't actually literals; they are applications of the `fromInteger` function.  This is true even in patterns; therefore they can't actually be matched as patterns but are translated into guards.

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