I'm sorry for asking such a basic question here, but I'm getting a syntax error when trying to compile the following code,
let sum_of_squares_of_two_largest x y z =
let a :: b :: _ = List.sort (fun x y -> -(compare x y)) [x; y; z] in
a * a + b * b;
let rec factorial n =
if n = 0 then 1 else n * factorial (n - 1);
let e_term n = 1.0 /. float_of_int (factorial n);
let rec e_approximation n =
if n = 0 then (e_term 0) else (e_term n) +. (e_approximation (n - 1));
let rec is_even x = if x = 0 then true else is_odd (x - 1);
and is_odd x = not (is_even x);
let rec f_rec n =
if n < 3 then n else f_rec(n - 1) + 2 * f_rec(n - 2) + 3 * f_rec(n - 3);
The uninformative compiler tells me there is syntax error on line 19, which is the last line of the file.
File "source.ml", line 19, characters 0-0:
Error: Syntax error
That is, line 19 is a blank line, only with a new-line character.
I can "fix" this syntax error by adding ;; at the end of each function definition instead of the ;.
Am I missing a semicolon somewhere?
As has been pointed out in the comments, ; is not a statement terminator like in many other (Algol-inspired) languages, but a sequence operator. It takes two values, throws away the first (but warns if it is not unit) and returns the second. a; b is therefore roughly equivalent to let _ = a in b.
You say you would have expected the error to say something like ';' is missing the second operand., but the thing is: it doesn't. Not until it reaches the end of the file (at which point the error message certainly could have been more intelligent, but probably not very accurate anyway).
What follows each ; in your code looks like a perfectly valid expression that might yield a value. For example, if let rec factorial n = ...; had been let rec factorial n = ... in 2 The value of the expression would have been 2. The problem here, from the compiler's point of view, is that it runs out of file before the expression is finished.
As you've also found out, ;; actually is a statement terminator. Or perhaps a toplevel definition terminator is a better term (I don't know if it even has an official name). It's used in the REPL to terminate input, but isn't needed in normal OCaml code unless you mix toplevel definitions and expressions, which you never should.
;; can still be useful for beginners as a "bulkhead", however. If you put just one ;; in place of a ; in the middle of your file, you'll find the compiler now points the error location there. That's because ;; terminates the definition without the preceding expression being complete. So you now know there's an error before that. (Actually in the preceding expression, but since your entire file is one single expression, "before that" is really the best we can do).
Related
I wrote a macro to make code cleaner and clearer by using error message templates stored in a dict. In this example it inserts name of the function, see discourse and #__FUNCTION__, that throws the error.
exception = Dict(:foo => ("bar ~", ArgumentError))
macro ⛔(id)
msg, type = exception[id]
quote
msg = replace($msg, "~" => StackTraces.stacktrace()[1].func)
throw($type(msg))
end
end
This works fine in f that uses positional arguments,
f(x, p) = p < 5 ? x : #⛔ foo
f(1, 10)
#> ERROR: ArgumentError: bar f
but displays #..# when a keyword argument is used.
g(x; p = 3) = p < 5 ? x : #⛔ foo
g(1, p = 10)
#> ERROR: ArgumentError: bar #g#N
Here N is the nth expression evaluated in the session. Redefining g with the same syntax increments this number.
I am stuck after I couldn't spot a difference between the code produced by #⛔.
#macroexpand function f(x, p) p < 5 ? x : #⛔ foo end
#macroexpand function g(x; p = 3) p < 5 ? x : #⛔ foo end
Question
What is happening in g that is different from f?
That is the name of the method the expanded code is called in -- definitions with keyword arguments are lowered to dispatch helper methods called "keyword sorters": https://docs.julialang.org/en/v1/devdocs/functions/#Keyword-arguments. These make keywords arguments use dispatch (internally) and compilation just as other functions, instead of just dictionary lookup as e.g. in Python.
I don't think you can do what you want easily in that case, as the conversion process always happens. Taking a previous stack frame would work, but then you have to know whether you are within a keyword method beforehand.
Maybe the following approach works: pattern match the method name (#<f>#<N>), and then check whether it is an actual kwsorter method of any method of f. If so, proceed with f.
I'm trying to write a function that creates a list of powers of a given number.
An example:
powList(2,5) = [0, 2, 4, 8, 16, 32]
Here is the code I already wrote:
let rec powList (x,n) =
if n = 0 then []
else (let a = let rec power (x, n) =
if n = 0 then 1
else x * power (x, n-1) in a) ::: powList(x, n-1);;
and this is the problem I get
Line 5, characters 31-32:
5 | else x * power (x, n-1) in a) ::: powList(x, n-1);;
^
Error: Syntax error: operator expected.
I'm only beginning to code in OCaml so I would be grateful for any help
Few generic remarks:
Don't define (recursive) functions inside recursive functions. Defining helper function in advance will make your code more readable and leave less room for syntax errors. For instance, lifting the definition of power outside of powList give you
let rec power x n =
if n = 0 then 1
else x * power x (n-1)
let rec powList (x,n) =
if n = 0 then []
else ( power ??? :: powList(x, n-1));;
Second, adding an element to a list is done with :: and not :::
Third, functions are generally curried in idiomatic OCaml code
let rec powList x n = ...
rather than
let rec powList (x,n)
because this opens more avenue for composition.
Fourth, your implementation is inefficient because it recomputes x^n at every step without fast exponentation. Consequently, your code ends up computing n * (n+1)/2 multiplications. Using fast exponentiation would reduce the number of multiplication to O(n log n). However the simpler version of using x^(n-1) to computes x yield only n multiplication:
let pow_list x n =
let rec pow_list x_power_n n =
if n = 0 then ...
else ...
in
pow_list ...
I'm learning OCaml, and the docs and books I'm reading aren't very clear on some things.
In simple words, what are the differences between
-10
and
~-10
To me they seem the same. I've encountered other resources trying to explain the differences, but they seem to explain in terms that I'm not yet familiar with, as the only thing I know so far are variables.
In fact, - is a binary operator, so some expression can be ambigous : f 10 -20 is treated as (f 10) - 20. For example, let's imagine this dummy function:
let f x y = (x, y)
If I want produce the tuple (10, -20) I naïvely would write something like that f 10 -20 which leads me to the following error:
# f 10 -20;;
Error: This expression has type 'a -> int * 'a
but an expression was expected of type int
because the expression is evaluated as (f 10) - 20 (so a substract over a function!!) so you can write the expression like this: f 10 (-20), which is valid or f 10 ~-20 since ~- (and ~+ and ~-. ~+. for floats) are unary operators, the predecense is properly respected.
It is easier to start by looking at how user-defined unary (~-) operators work.
type quaternion = { r:float; i:float; j:float; k:float }
let zero = { r = 0.; i = 0.; j = 0.; k = 0. }
let i = { zero with i = 1. }
let (~-) q = { r = -.q.r; i = -.q.i; j = -. q.j; k = -. q.k }
In this situation, the unary operator - (and +) is a shorthand for ~- (and ~+) when the parsing is unambiguous. For example, defining -i with
let mi = -i
works because this - could not be the binary operator -.
Nevertheless, the binary operator has a higher priority than the unary - thus
let wrong = Fun.id -i
is read as
let wrong = (Fun.id) - (i)
In this context, I can either use the full form ~-
let ok' = Fun.id ~-i
or add some parenthesis
let ok'' = Fun.id (-i)
Going back to type with literals (e.g integers, or floats), for those types, the unary + and - symbol can be part of the literal itself (e.g -10) and not an operator. For instance redefining ~- and ~+ does not change the meaning of the integer literals in
let (~+) = ()
let (~-) = ()
let really = -10
let positively_real = +10
This can be "used" to create some quite obfuscated expression:
let (~+) r = { zero with r }
let (+) x y = { r = x.r +. y.r; i = x.i +. y.i; k = x.k +. x.k; j =x.k +. y.j }
let ( *. ) s q = { r = s *. q.r; i = s *. q.i; j = s *. q.j; k = s *. q.k }
let this_is_valid x = +x + +10. *. i
OCaml has two kinds of operators - prefix and infix. The prefix operators precede expressions and infix occur in between the two expressions, e.g., in !foo we have the prefix operator ! coming before the expression foo and in 2 + 3 we have the infix operator + between expressions 2 and 3.
Operators are like normal functions except that they have a different syntax for application (aka calling), whilst functions are applied to an arbitrary number of arguments using a simple syntax of juxtaposition, e.g., f x1 x2 x3 x41, operators can have only one (prefix) or two (infix) arguments. Prefix operators are very close to normal functions, cf., f x and !x, but they have higher precedence (bind tighter) than normal function application. Contrary, the infix operators, since they are put between two expressions, enable a more natural syntax, e.g., x + y vs. (+) x y, but have lower precedence (bind less tight) than normal function application. Moreover, they enable chaining several operators in a row, e.g., x + y + z is interpreted as (x + y) + z, which is much more readable than add (add (x y) z).
Operators in OCaml distinguished purely syntactically. It means that the kind of an operator is fully defined by the first character of that operator, not by a special compiler directive, like in some other languages (i.e., there is not infix + directive in OCaml). If an operator starts with the prefix-symbol sequence, e.g., !, ?#, ~%, it is considered as prefix and if it starts with an infix-symbol then it is, correspondingly, an infix operator.
The - and -. operators are treated specially and can appear both as prefix and infix. E.g., in 1 - -2 we have - occurring both in the infix and prefix positions. However, it is only possible to disambiguate between the infix and the prefix versions of the - (and -.) operators when they occur together with other operators (infix or prefix), but when we have a general expression, the - operator is treated as infix. E.g., max 0 -1 is interpreted as (max 0) - 1 (remember that operator has lower precedence than function application, therefore when they two appear with no parentheses then functions are applied first and operators after that). Another example, Some -1, which is interpreted as Some - 1, not as Some (-1). To disambiguate such code, you can use either the parentheses, e.g., max 0 (-1), or the prefix-only versions, e.g, Some ~-1 or max 0 ~-1.
As a matter of personal style, I actually prefer parentheses as it is usually hard to keep these rules in mind when you read the code.
1) Purists will say that functions in OCaml can have only one argument and f x1 x2 x3 x4 is just ((f x1) x2) x3) x4, which would be a totally correct statement, but a little bit irrelevant to the current discussion.
What's the problem with the code below? I'am struggling a lot with knowing when to use ';' or ';;' or use begin end in OCaml. Here i need to read some edges and insert into graph but i need to link this with rest of program so that it can use g(a graph) with all the edges.
When i do this it says Error: Syntax error
let i = ref n in
while !i > 0 do
(
let pair = read_edge Scanning.stdin in
let g = insert_edge (fst pair) (snd pair) g in
i := !i - 1
)
done in (* giving error in this line *)
let rec do_stuff l =
match l with
| [] -> ()
| h::t -> print_int h;do_stuff t in
( * more functions)
in is part of the syntax of let (it's "let" pattern "in" expression). The syntax of a while loop is simply "while" expression "do" expression "done", so there's no in in there.
To make your code compile you can replace in with a ;, so it executes the while loop followed by the let expression after it. However it seems strange to me that you'd have a while loop between a bunch of function definitions.
Also note that in the loop you do let g = ..., but then you're not using g anywhere in the let's body. So that binding accomplishes nothing.
The following method determines how many numbers can be added up starting from the beginning of the list without adding up to 4:
number_before_Reaching_sum (4, [1,2,3,4,6]);
should return : val it = 2 : int
fun number_before_reaching_sum (sum : int * int list) =
let val len_orig_list = length (#2 sum)
in fun num_bef_reach_sum (sum) =
if #1 sum <= 0
then len_orig_list - (length (#2 sum)) - 1
else num_bef_reach_sum (#1 sum - hd (#2 sum), tl (#2 sum))
end
syntax error: inserting LOCAL
syntax error found at EOF
I can't seem to find the errors in this code. I have some experience with Python, but just starting to learn sml. I'm loving it, but I don't understand all the error messages. I have really spent hours on this, but I think I don't know enough to solve my problem. I tried exchanging let with local, but i still got a syntax error (equalop). I think that the function between in and end is an expression and not a declaration. But I would appreciate any comments on this. If you come up with alternative code, it would be great if you did it without using more advanced features, since I'm just trying to get the basics down :-)
You probably meant this:
fun number_before_reaching_sum (sum : int * int list) =
let
val len_orig_list = length (#2 sum)
fun num_bef_reach_sum (sum) =
if #1 sum <= 0
then len_orig_list - (length (#2 sum)) - 1
else num_bef_reach_sum (#1 sum - hd (#2 sum), tl (#2 sum))
in
num_bef_reach_sum (sum)
end
With let … in … end, the part between let and in is for the local definitions; the part between in and end is for the expression which will be the evaluation of the let … in … end expression (this construct is indeed an expression).
Think of let … in … end as a possibly complex expression. You hoist parts of the expression as definitions, then rewrite the complex expression using references to these definitions. This help write shorter expressions by folding some of its sub‑expressions. This construct is also required when recursion is needed (a recursion requires a defining name).
Another way to understand it, is as the application of an anonymous function whose arguments bindings are these definitions.
Ex.
let
val x = 1
val y = 2
in
x + y
end
is the same as writing
(fn (x, y) => x + y) (1, 2)
which is the same as writing
1 + 2
You erroneously put a definition at the place of the expression the whole evaluates to.
(note I did not check the function's logic, as the question was about syntax)