I want to make a recursive function that sums the integers between two values. I'm doing:
let rec sum_between x y =
if x>y then sum_between y x else
if x=y then x else x + sum_between x+1 y ;;
But I get the error: This expression has type int -> int
but an expression was expected of type int
What am I doing wrong?
Function application has high precedence in OCaml. You need to parenthesize an expression when it's an argument to a function.
Your code
sum_between x+1 y
is parsed like this:
(sum_between x) + (1 y)
You need parentheses:
sum_between (x + 1) y
(Same answer as Edgar Aroutiounian but more helpful detail I hope.)
Related
I am having trouble understanding a method to implement a power function in SML using only iteration and multiplication.
my iteration is the following:
fun iterate 0 f x = x
| iterate n f x = iterate (n-1) f (f x);
while my multiplication is basically iterating recursively
fun multiply 0 f = 0
| multiply f x = iterate x (fn x => x + 1) (multiply x (f-1));
Power function would basically be an iteration of the multiplication of the same base but I don't know which value to decrement
power n f = iterate (mult n n) (fn x => x + 1) (power (n) (f-1))
which is definately wrong
power n f = iterate (mult n n) (fn x => x + 1) (power (n) (f-1))
So, when it comes to naming, I might definitely write power x y or power i j or power x n or some such, since x, y, i, j or n look like they're numbers or integers, whereas f looks like it's a function. So right off the bat you have:
fun power x y = iterate (...a...) (...b...) (...c...)
As for what goes into each of these three parts, ...a..., ...b... and ...c...:
a. The thing iterate calls n, which is the number of times to iterate.
b. The thing iterate calls f, which is the function to apply each time.
c. The thing iterate calls x, which is what is applied each time.
As elaborated on in How to make a multiplication function using just addition function and iterate function in SML, there is no point in making power call itself; the point of using iterate is to hand over recursion to this list-combinator rather than use explicit recursion (where power has a reference to itself somewhere in its definition).
I am new to haskell code. I tried to compute the sum of squares of negative integer in a list using foldr high order.
sumsq :: Int -> Int
sumsq n = foldr op 0 [1..n]
where op x y = x*x + y
Please help to explain each line of code and give any solution if error in this code
When using "where", important to follow the indentation rule.
Here lambda will be appropriate
sumsq n = foldr (\x y -> x*x + y) 0 [1..n]
I am new to Standard ML. I am trying to compute x squared i, where x is a real and i is an non-negative integer. The function should take two parameters, x and i
Here is what I have so far:
fun square x i = if (i<0) then 1 else x*i;
The error that I am getting is that the case object and rules do not agree
The unary negation operator in SML is not - as it is in most languages, but instead ~. That is likely what is causing the specific error you cite.
That said, there are some other issues with this code. L is not bound in the example you post for instance.
I think you may want your function to look more like
fun square (x : real) 0 = 1
| square x i = x * (square x (i - 1))
You'll want to recurse in order to compute the square.
i wrote a pow function in ocaml and by mistake a wrote this
let rec pow x y acc = if y = 1 then acc else pow x (y-1) x*x;;
this function can be called with any value of acc and output the correct answer, but this must output x^2 .
this is the correct functon :
let rec pow x y acc = if y = 0 then acc else pow x (y-1) x*acc;;
and call this function with the value of acc of 1.
my question is why first function give the correct output ?
The key is that function application has higher precedence than binary operators. In the case at hand, the else term is equivalent to (pow x (y-1) x) * x, not pow x (y-1) (x*x).
I am trying to calculate the square root of a number by Ocaml, this is my code :
let isclose x y = abs_float(x -. y)<0.001;;
let average x y = (0.5*.x)+.(0.5*.y);;
let rec guess x y = if isclose y (x /. y) then y else guess x (average y x/.y);;
let sqr x = guess x 1.;;
then, typing
sqr 1.;;
gives 1 as expected, but typing
sqr 2.;;
lasts undefinitely.
Can one help with my error (I tested the algo in Python and it works as expected).
Thanks for help
You want this:
let rec guess x y =
if isclose y (x /. y) then y else guess x (average y (x/.y))
Note the extra parentheses.
The meaning of
average y x /. y
is
(average y x) /. y
Whereas you want:
average y (x /. y)