Re: [Haskell-beginners] Compiler can't deduce Bool as instance of ToField

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Bryan Vicknair: postgresql-simple declares Bool as an instance of the ToField class. The compiler can't deduce that given this simple code however:

import Database.PostgreSQL.Simple.ToField (ToField(..))

foo :: (ToField a) => a foo = True

It fails with this error:

Db.hs:64:7: Could not deduce (a ~ Bool) from the context (ToField a) bound by the type signature for foo :: ToField a => a at Db.hs:63:8-23 `a' is a rigid type variable bound by the type signature for foo :: ToField a => a at Db.hs:63:8 In the expression: True In an equation for `foo': foo = True Failed, modules loaded: none.

What am I missing?

Oliver Charles You have declared that foo is *any* type that has a ToField instance, allowing the caller of foo to determine the type at their will. However, your implementation of foo is more specific and requires a is actually Bool and nothing else. On 14 Aug 2013 10:24, "Bryan Vicknair" wrote:

I see. I was confused, because the following works: bar :: (Num a) => a bar = 5 I guess it is because the literal '5' could be an Int or a Float, and the type system doesn't know. When I saw that the following does not work...:: bar :: (Show a) => a bar = False ...it made a bit more sense. It seems that if a literal can be considered part of a typeclass in more than one way, you can declare that literal to be of that typeclass. However, if there is only one way for a literal to belong to a typeclass, then you can't. Are there other literals (besides (Num a) => [a]) that don't have a concrete type like literal numbers? If I understand correctly, -XOverloadedStrings turns all literal strings into IsString instead of [Char], so that seems to be another case. I don't have a need for this, but now I'm curious: is there a way to declare that all literal numbers in a module are of a certain concrete type? Something like -XAllNumbersAreInt8?