> f :: a -> b -> c is a function that takes an a, a b, and returns a c.
>
> g :: (a -> b) -> c takes one argument, which is expected to be a
> function from a to b. g returns a c.
>
> That stuff I mentioned before about variable binding and function
> application still applies. We can show that f and g have "isomorphic"
> types. But they are conceptually different types.
Except that f and g are not isomorphic. In fact, there exists no defined
fuction g :: (a -> b) -> c
(what type would (g id) be?)
Perhaps you meant g :: a -> (b -> c)?
Alexander Solla wrote:
On Jan 26, 2010, at 8:11 AM, Xingzhi Pan wrote:
I'm a little confused with the type of step here. Can it be
considered as taking 2 or 3 arguments and then the compiler has to
infer to decide?
The compiler is going to build up a sequence of functions. Consider
the following (mathematical) function:
f(x, y, z) = x^2 + y^2 + z^2.
This is a function in three arguments. But if you bind any of them
(say, x) to a value x', you end up with a function g(y,z) =
f(x',y,z). This is a function in two arguments. Bind another
variable, and you get another function, with one less argument. You
need to bind all the variables in order to compute f for a point.
Say if I, as a code reader, meet such a function
defined with three formal parameters, how can I draw the conclusion of
its type (and it takes 2 arguments actually)?
You can derive this from the syntactic properties of types. Count the
number of arrows that are not "in parentheses". That's how many
arguments the function takes.
f :: a -> b -> c is a function that takes an a, a b, and returns a c.
g :: (a -> b) -> c takes one argument, which is expected to be a
function from a to b. g returns a c.
That stuff I mentioned before about variable binding and function
application still applies. We can show that f and g have "isomorphic"
types. But they are conceptually different types.
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