Rather than bailing with an instance error for both Applicative and Num, which is the technically the right way, and the way that it used to be in the dark ages of ghci. Instead it chooses types which are probably what you wanted. In this case it defaults f to IO and a to Int, and then runs it. You can read more about type defaulting in ghci here: https://downloads.haskell.org/~ghc/latest/docs/html/users_guide/ghci.html On Thu, Dec 22, 2016 at 10:18 AM, Ovidiu Deac wrote:

My understanding is that x can take any form required for type-inference. That's fine but what is the "default" structure if you don't specify any?

On Thu, Dec 22, 2016 at 5:13 PM, Imants Cekusins wrote:

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What is f here?

anything Applicative:

Prelude> let a1 = pure 1 Prelude> let a2 = pure 1

Prelude> (a1::Maybe Int) == a2 True Prelude> (a1::Maybe Float) == a2 True Prelude> (a1::Either String Float) == a2 True

​

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what is the "default" structure if you don't specify any

similar to: display::Show a => a -> String display = show fa::(Applicative f, Num a) => f a fa = pure 1 f and a are *bounded* by Applicative and Num, so to say. No default. Or, it is typed however type is a bit broader than *

​ In this case it defaults f to IO and a to Int,

does it though? Prelude> let a1 = pure 1 Prelude> let a2 = pure 1 Prelude> a1 == a2 <interactive>:20:1: error: • Ambiguous type variable ‘a0’ arising from a use of ‘a1’ prevents the constraint ‘(Num a0)’ from being solved.

I get a Num ambiguous constraint error as well, but it is merely not giving you the rest of the errors. Prelude> let a1 = pure (1::Int) Prelude> let a2 = pure (1::Int) Prelude> a1 == a2 • Ambiguous type variable ‘f0’ arising from a use of ‘a1’ prevents the constraint ‘(Applicative f0)’ from being solved. Furthermore you can't compare an IO a with an IO a, so it will still cause an error. But if you go: Prelude>let a1 = 1 Prelude>let a2 = 1 Prelude>a1 == a2 True Then it is totally fine with that. On Thu, Dec 22, 2016 at 10:38 AM, Imants Cekusins wrote:

...

what is the "default" structure if you don't specify any

similar to:

display::Show a => a -> String display = show

fa::(Applicative f, Num a) => f a fa = pure 1

f and a are *bounded* by Applicative and Num, so to say. No default. Or, it is typed however type is a bit broader than *

...

​ In this case it defaults f to IO and a to Int,

does it though?

Prelude> let a1 = pure 1 Prelude> let a2 = pure 1 Prelude> a1 == a2

<interactive>:20:1: error: • Ambiguous type variable ‘a0’ arising from a use of ‘a1’ prevents the constraint ‘(Num a0)’ from being solved.

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