The point is that in
f::a
you are saying that f is of an undefined type a. But in
f=1
you're saying that a must be a type that supports the value 1, and not all do (eg strings, booleans).
Therefore you have to tell the compiler that you are constraining a to allow it to take the value 1, hence the constraint Num a, which does just that.
The error message is telling you just this: it can't find a constraint that tells it that a is in the class Num, so it is complaining because what you have written could fail if you used f in a context where it was inferred to have a non numerical type. Constraints allow the type-inference engine to spot such errors, saving you from terrible grief in debugging.
Sent from my iPhone
On 29 Jul 2011, at 00:42, Jake Penton
Yikes.
I have been doing a fair bit of productive programming in Haskell, thinking that I am making a bit of progress. Then I hit something that is apparently *really simple* that I do not understand at all. How discouraging.
Here is the code that makes me realize I don't understand types or type inference very well at all yet:
f :: a f = 1
When I try to load the above, ghci gives me:
No instance for (Num a) arising from the literal `2' In the expression: 2 In an equation for `f': f = 2 Failed, modules loaded: none.
Ok, so then I try:
g:: (Num a) => a g = 2
This compiles.
Why? I mean, why is the first example (defining f) wrong, but the second example (defining g) ok?
A slight variation on this is:
h:: a h = 'a'
to which ghci replies:
Couldn't match type `a' with `Char' `a' is a rigid type variable bound by the type signature for c :: a at /Users/David/Project/EoP/ch04/weak.hs:114:1 In the expression: 'a' In an equation for `c': c = 'a'
This last example is probably the most basic one which I need to understand. But, why is the problem apparently a different one than in the definition of "f" above?
Of course, I cannot think of a reason to actually define things as shown above under ordinary circumstances. The code above is just boiled down to the simplest case I could find to illustrate my confusion.
I guess I interpret "f::a" to mean "f is some (any) type a". So why can't it be whatever "1" is, which I suppose is Integer. What is the type system looking for? And why does the constraint (Num a) make things ok?
Please point me in the direction of any reading I should do to clear up my confusion.
TIA.
- Jake - _______________________________________________ Beginners mailing list Beginners@haskell.org http://www.haskell.org/mailman/listinfo/beginners