[Haskell-beginners] binding vs. parameter passing
Ramnath R Iyer
rri at silentyak.com
Tue Jul 23 03:44:47 UTC 2019
This answer is likely incomplete at best and wrong at worst, but I'll give
it a shot.
When you have this type declaration:
i :: a
It means that you are declaring the symbol i that can be slotted into any
location that requires an a. It is a task for the compiler to process the
overall program specification and infer that a has a particular concrete
type. When the program is specified though, the value i has a generic type
with no bounds, and the only way to *construct* such a value is to supply a
value of type a from somewhere else.
The problem here is not that you can't *bind* 1 to i, but that you cannot
*construct* any value of type a.
A direct analog of this would be - not the application of the id function
as you called out - but that you cannot define a function with
the following signature, not including bottom:
f :: a -> b
The reason is that you can't supply a value of type b.
I have some experience with Java, and it has a similar constraint that
generic types cannot be instantiated.
On Mon, Jul 22, 2019 at 7:30 PM James Jones <jejones3141 at gmail.com> wrote:
> One problem in *Programming Haskell from First Principles* confuses me.
> (So far.)
> It's one where you start with the declaration and binding
> i :: Num a => a
> i = 1
> (which of course works because Num a => a is the type of 1) and then
> change the declaration to
> i :: a
> and first try to predict and then see what happens. It fails, with ghci
> suggesting that you put the Num a => back. That seemed reasonable at
> first, but then I considered this:
> - id has type a -> a, and if you try to evaluate id 1 it works without
> - In all the work I've done on compilers, parameter passing has
> effectively been assignment of actual parameters to the corresponding
> formal parameters. In Haskell, that might mean passing along the
> appropriate thunk, but the principle is the same, isn't it?
> So, if I can't bind 1 to i which is declared to have type a, why can I
> successfully pass 1 to id?
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> Beginners at haskell.org
Ramnath R Iyer
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