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
Related
I'm reading this OCaml file and it contains the following:
type z = Z of z
This looks like z is infinitely recursive. How is it useful and how can I even construct such a type?
I don't think this type is particularly useful, except possibly as a test case in type theory.
You can construct values of the type like this:
# let rec x = Z x;;
val x : z = Z <cycle>
# let rec q = Z (Z q);;
val q : z = Z (Z <cycle>)
Once you have a value of the type you can readily construct other values of course:
# let y = Z (Z x);;
val y : z = Z (Z (Z <cycle>))
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've written a function that calculates a value of x, of a polynomial made from a list of reals.
infixr 5 ^^;
fun (x:real) ^^ 0 = 1.0
| (x:real) ^^ n = x*(x^^(n-1));
fun poly [] (x:real) = 0.0
| poly (hd::tl) (x:real) = hd*(x^^(length tl)) + poly tl x;
This code all works perfectly fine, and I'm quite proud of it.
I have managed to create polynomial functions using partial application:
> fun f x = poly [3.0,2.0,1.0] x;
val f = fn : real -> real
> f 2.0;
val it = 17.0 : real
Creating the mathetmatical function:f(x) = 3*x^2 + 2*x + 1
This is all fine, but I want to be able to construct a function by this method:
fun f x = polyGen [1.0,2.0,3.0];
And it will give me an equivalent function to the one above.
Is this possible?
I know it seems trivial, I could just put an x there as I did before and get on with my life. But I'm just curious on how someone would get around this problem!
Thanks in advance, Ciaran
EDIT:
fun polyGen L = let fun poly [] x = 0.0
| poly (hd::tl) x = hd + x*(poly tl x);
in fn x => poly L x end;
Lovely!
If I understand your question correctly, then you don't need to define anything else at all. With the function poly that you have you can already do
val f = poly [3.0, 2.0, 1.0]
which defines f as a function of type real -> real.
I'm doing some homework but I've been stuck for hours on something.
I'm sure it's really trivial but I still can't wrap my head around it after digging through the all documentation available.
Can anybody give me a hand?
Basically, the exercise in OCaml programming asks to define the function x^n with the exponentiation by squaring algorithm.
I've looked at the solution:
let rec exp x = function
0 -> 1
| n when n mod 2 = 0 -> let y = exp x (n/2) in y*y
| n when n mod 2 <> 0 -> let y = exp x ((n-1)/2) in y*y*x
;;
What I don't understand in particular is how the parameter n can be omitted from the fun statement and why should it be used as a variable for a match with x, which has no apparent link with the definition of exponentiation by squaring.
Here's how I would do it:
let rec exp x n = match n with
0 -> 1
| n when (n mod 2) = 1 -> (exp x ((n-1)/2)) * (exp x ((n-1)/2)) * x
| n when (n mod 2) = 0 -> (exp x (n/2)) * (exp x (n/2))
;;
Your version is syntaxically correct, yields a good answer, but is long to execute.
In your code, exp is called recursively twice, thus yielding twice as much computation, each call yielding itself twice as much computation, etc. down to n=0. In the solution, exp is called only once, the result is storred in the variable y, then y is squared.
Now, about the syntax,
let f n = match n with
| 0 -> 0
| foo -> foo-1
is equivalent to:
let f = function
| 0 -> 0
| foo -> foo-1
The line let rec exp x = function is the begging of a function that takes two arguments: x, and an unnammed argument used in the pattern matching. In the pattern matching, the line
| n when n mod 2 = 0 ->
names this argument n. Not that a different name could be used in each case of the pattern matching (even if that would be less clear):
| n when n mod 2 = 0 -> let y = exp x (n/2) in y*y
| p when p mod 2 <> 0 -> let y = exp x ((p-1)/2) in y*y*x
The keyword "function" is not a syntaxic sugar for
match x with
but for
fun x -> match x with
thus
let rec exp x = function
could be replaced by
let rec exp x = fun y -> match y with
which is of course equivalent with your solution
let rec exp x y = match y with
Note that i wrote "y" and not "n" to avoid confusion. The n variable introduced after the match is a new variable, which is only related to the function parameter because it match it. For instance, instead of
let y = x in ...
you could write :
match x with y -> ...
In this match expression, the "y" expression is the "pattern" matched. And like any pattern, it binds its variables (here y) with the value matched. (here the value of x) And like any pattern, the variables in the pattern are new variables, which may shadow previously defined variables. In your code :
let rec exp x n = match n with
0 -> 1
| n when (n mod 2) = 1 -> (exp x ((n-1)/2)) * (exp x ((n-1)/2)) * x
| n when (n mod 2) = 0 -> (exp x (n/2)) * (exp x (n/2))
;;
the variable n in the two cases shadow the parameter n. This isn't a problem, though, since the two variable with the same name have the same value.