An exercise questions asks for an explanation as to why this won't work. Obviously from running the code I see that it doesn't, but I don't see in this case why. The error doesn't clarify much!
# let (+) x y z = x + y + z in 5 + 6 7;;
Error: This expression has type int
This is not a function; it cannot be applied.
Thanks!
Lets go step-by-step. Fire a REPL and type:
# let (+) x y z = x + y + z;;
val ( + ) : int -> int -> int -> int = <fun>
We can interpret this int -> int -> int -> int as an infix + operator that takes two ints and returns an int -> int function.
Lets check that:
# let f = 5+6;;
val f : int -> int = <fun>
# f 7;;
- : int = 18
That's every step of your intended program working.
The issue with your code is that this doesn't work:
# 5+6 7;;
Error: This expression has type int
This is not a function; it cannot be applied.
That happens because function application has a precedence over + operator. (In fact, function application has the strongest precedence in OCaml.) So adding the brackets, fixes it (you'll need to restart the toplevel):
# let (+) x y z = x + y + z in (5+6) 7;;
- : int = 18
Related
Is it possible to declare multiple variable bindings in one line in SML? For example, I have the following:
let
val m1 = [1]
val m2 = [2]
val m3 = [3]
in
{...}
end
I would like to condense this down to something like
let
val m1 = [1], m2 = [2], m3 = [3]
in
{...}
end
This syntax doesn't work, but is there a way to declare multiple variable bindings in one line like this?
Here are two ways:
- let val x = 1 val y = 5 in x + y end;
val it = 6 : int
- let val x = 1 and y = 5 in x + y end;
val it = 6 : int
I personally find the slight abuse of and more readable.
However, I think the "destructuring bind" method is more common than either of these, since it's also more generally useful.
- let val (x, y) = (1,5) in x + y end;
val it = 6 : int
- fun f x = (x, x + 2, x + 3);
val f = fn : int -> int * int * int
- let val (x, y, z) = f 3 in x + z end;
val it = 9 : int
You could create a tuple and immediately destructure it.
let
val (m1, m2, m3) = ([1], [2], [3])
in
...
end
When f(x-1) is called, is it calling f(x) = x+10 or f(x) = if ...
Is this a tail recursion?
How should I rewrite it using static / dynamic allocation?
let fun f(x) = x + 10
in
let fun f(x) = if x < 1 then 0 else f(x-1)
in f(3)
end
end
Before addressing your questions, here are some observations about your code:
There are two functions f, one inside the other. They're different from one another.
To lessen this confusion you can rename the inner function to g:
let fun f(x) = x + 10
in
let fun g(x) = if x < 1 then 0 else g(x-1)
in g(3)
end
end
This clears up which function calls which by the following rules: The outer f is defined inside the outer in-end, but is immediately shadowed by the inner f. So any reference to f on the right-hand side of the inner fun f(x) = if ... is shadowed because fun enables self-recursion. And any reference to f within the inner in-end is shadowed
In the following tangential example the right-hand side of an inner declaration f does not shadow the outer f if we were using val rather than fun:
let fun f(x) = if (x mod 2 = 0) then x - 10 else x + 10
in
let val f = fn x => f(x + 2) * 2
in f(3)
end
end
If the inner f is renamed to g in this second piece of code, it'd look like:
let fun f(x) = if (x mod 2 = 0) then x - 10 else x + 10
in
let val g = fn x => f(x + 2) * 2
in g(3)
end
end
The important bit is that the f(x + 2) part was not rewritten into g(x + 2) because val means that references to f are outer fs, not the f being defined, because a val is not a self-recursive definition. So any reference to an f within that definition would have to depend on it being available in the outer scope.
But the g(3) bit is rewritten because between in-end, the inner f (now g) is shadowing. So whether it's a fun or a val does not matter with respect to the shadowing of let-in-end.
(There are some more details wrt. val rec and the exact scope of a let val f = ... that I haven't elaborated on.)
As for your questions,
You should be able to answer this now. A nice way to provide the answer is 1) rename the inner function for clarity, 2) evaluate the code by hand using substitution (one rewrite per line, ~> denoting a rewrite, so I don't mean an SML operator here).
Here's an example of how it'd look with my second example (not your code):
g(3)
~> (fn x => f(x + 2) * 2)(3)
~> f(3 + 2) * 2
~> f(5) * 2
~> (if (5 mod 2 = 0) then 5 - 10 else 5 + 10) * 2
~> (if (1 = 0) then 5 - 10 else 5 + 10) * 2
~> (5 + 10) * 2
~> 15 * 2
~> 30
Your evaluation by hand would look different and possibly conclude differently.
What is tail recursion? Provide a definition and ask if your code satisfies that definition.
I'm not sure what you mean by rewriting it using static / dynamic allocation. You'll have to elaborate.
This question already has answers here:
Value of bindings in SML?
(2 answers)
Closed 6 years ago.
Could someone please help. I don't get the sequence of evaluation here and how we got values of "ans". e.g. in the first example there's no value of y and I'm not sure whether this returns a pair or calls x ! (fn y => y x). It would be very helpful if you can Trace each expression.
val x = 1
val f = (fn y => y x)
val x = 7
val g = (fn y => x - y)
val ans = f g
val ans = 6 : int
=====================================
fun f p =
let
val x = 3
val y = 4
val (z,w) = p
in
(z (w y)) + x
end
val x = 1
val y = 2
val ans = f((fn z => x + z), (fn x => x + x + 0))
val ans = 12 : int
There are a few things which help make problems like this much clearer
when trying understand an alien function Lexical scoping works.
add in types to the parameters and return values without modifying the program, the compiler will tell you if you get it wrong...
replace anonymous functions with named ones.
rename variable bindings that have the same names but refer to different lexical scope.
remove variable bindings that only get used once.
binding a value to a name does not actually perform any computation,
so is merely for the benefit of the reader, if it is not doing that job
it merely serves to obfuscate, then by all means remove it.
fun f (y1 : int -> 'a) = y1 1 : 'a;
fun g (y2 : int) = 7 - y2 : int;
val ans : int = f g;
so g is given as a parameter to f, f calls g giving it the parameter x having the value 1 making y2 = 1, which g subtracts 7 - 1 returning 6.
the return value of g is an int, thus f's 'a type when g is applied to it is an int.
for the 2nd one clean it up a bit, I pulled the anonymous fn's out into their own and named values and call f (foo, bar) to make it more readable...
fun f p =
let val x = 3
val y = 4
val (z, w) = p
in (z (w y)) + x end
fun foo z = z + 1;
fun bar x = x * 2;
val ans = f(foo, bar);
Finally, we can get rid of the let values which are only used once
and replace the (z,w) = p with just (z, w) as a parameter to the function which should be much easier to follow
fun f (z, w) = (z (w 4)) + 3
fun foo z = z + 1;
fun bar x = x * 2;
val ans = f(foo, bar);
val ans = ((4 * 2) + 1) + 3
I have a question why does OCaml behaves somewhat unusual. By defining the function
let abs_diff x y = abs(x - y);;
we get val abs_diff : int -> int -> int = <fun> now by defining as
let abs_diff x y =
fun x -> (fun y -> abs(x - y));;
val abs_diff : 'a -> 'b -> int -> int -> int = <fun>
now using another function called as
let dist_from3 = abs_diff 3;;
with the first definition it works perfectly but with the second one it does not work as expected. We get that it is
val dist_from3 : '_a -> int -> int -> int = <fun>
why is it behaving like that, and why are those two definition of the at first look same function different?
In your second definition you have two distinct appearances (bindings) of x and y. That's why there are four parameters in the result. This is what you want:
let abs_diff = fun x -> fun y -> abs (x - y)
(FWIW in actual practice I sometimes make this mistake, especially when using the function keyword.)
let (++) f g x = f (g x) in
let f x = x + 1 in
let g x = x * 2 in
(f++g) 1;;
Is the above expression correct?
It seems to me that the above code should be just like defining f++g x = 2 * x + 1. Am I correct?
Your implementation of function composition is correct, since :
(g ∘ f)(x) = g(f(x)) for all x in X
according to wikipedia
I get :
- : int = 3
in ocamlktop