I wanted to create a global variable called result that uses 5 string concatenations to create a string containing 9 times the string start, separated by commas.
I have two pieces of code, only the second one declares a global variable.
For some reason it's not registering easily in my brain... Is it just that i used a let in so result in the first piece of code is a local variable? Is there a more detailed explanation for this?
let start = "ab";;
let result = start ^ "," in
let result = result ^ result in
let result = result ^ result in
let result = result ^ result in
let result = result ^ start in
result;;
- : string = "ab,ab,ab,ab,ab,ab,ab,ab,ab"
let result =
let result = start ^ "," in
let result = result ^ result in
let result = result ^ result in
let result = result ^ result in
let result = result ^ start in
result;;
val result : string = "ab,ab,ab,ab,ab,ab,ab,ab,ab"
Let me to be a little bit boring person. There are no local and global variables in OCaml. This concept came from languages with different scoping rules. Also, the word "variable" itself should be taken with care. Its meaning was perverted by C-like languages. The original, mathematical, meaning of this word corresponds to a name of some mathematical object, that is used inside a formula, that represent a range of such values. In C-like languages, a variable is confused with the memory cell, that can change in time. So, to avoid the confusion let's use a more accurate terminology. Let's use word name instead of variable. Since, variables... sorry names are not memory cells, there is nothing to create. When you're using one of the let syntaxes, you're actually creating a binding, i.e., an association between a name and a value. The let <name> = <expr-1> in <expr-2> binds a value of the in the scope of the <expr-2> expression. The let <name> = <expr-1> in <expr-2> is by itself is also an expression, so, for example <expr-2> can also contain let ... in ... constructs inside, e.g.,
let a = 1 in
let b = a + 1 in
let c = b + 1 in
a + b + c
I especially, indented the code in non-idiomatic way to highlight the syntactic structure of the expression. OCaml also allows to use a name, that is already bound in the scope. The new binding will hide the existing one (that is not allowed in C, for example), e.g.,
let a = a + 1 in
let a = a + 1 in
let a = a + 1 in
a + a + a
Finally, the top-level (aka module level) let-binding (called definition in OCaml parlance), has the syntax: let <name> = <expr>, note that there is no in here. The definition binds the <name> to a result of the evaluation of <expr> in the lexical scope that extends form the point of definition to the end of the enclosing module. When you're implementing a module, you must use let <name> = <expr> to bind your code to names (you may omit name by using _). It is a little bit different from the interactive toplevel (interactive ocaml program), that actually accepts an expression, and evaluates it. For example,
let result = start ^ "," in
let result = result ^ result in
let result = result ^ result in
let result = result ^ result in
let result = result ^ start in
result
Is not a valid OCaml program (something that can be put into an ml file and compiled). Because it is an expression, not a module definition.
Is it just that i used a let in so result in the first piece of code is a local variable?
Pretty much. The syntax to define a global variable is let variable = expression without an in. The syntax to define a local variable is let variable = expression in expression which will define variable local to the expression after the in.
When you have let ... in, you're declaring a local variable. When you have just let by itself (at the top level of a module), you're declaring a global name of the module. (That is, a name that can be exported from the module.)
Your first example consists entirely of let ... in. So there is no top-level name declared.
Your second example has one let by itself, followed by several occurrences of let ... in. So it declares a top-level name result.
Related
The following code comes from Stylish F# 6: Crafting Elegant Functional Code for .NET 6 listing 9-13:
let randomByte =
let r = System.Random()
fun () ->
r.Next(0, 255) |> byte
// E.g. A3-52-31-D2-90-E6-6F-45-1C-3F-F2-9B-7F-58-34-44-
for _ in 0..15 do
printf "%X-" (randomByte())
printfn ""
The author states, "Although we call randomByte() multiple times, only one System.Random() instance is created."
I understand randomByte returns a function that does not create a System.Random() instance, but it seems to me multiple System.Random() instances would be created each time through the for-do-loop anyway.
I would appreciate an explanation of how multiple instances of System.Random() are not created in this case.
The key point is that randomByte is not a function. It's a value with some complex initialization logic. Like, for example, I could write:
let x = 5
Or I could write:
let x =
let fourtyTwo = 42
let thirtySeven = 37
fourtyTwo - thirtySeven
And these would be equivalent. Both declare a value named x and equal to 5. I hope you can see how the expression fourtyTwo - thirtySeven is evaluated only once, not every time somebody gets the value of x.
And so it works with randomByte too: it's a value with non-trivial initialization logic. During that value's initialization, first it creates an instance of System.Random, and then it creates an anonymous function that closes over that instance, and this anonymous function becomes the value of randomByte.
For a list, you can do pattern matching and iterate until the nth element, but for a tuple, how would you grab the nth element?
TL;DR; Stop trying to access directly the n-th element of a t-uple and use a record or an array as they allow random access.
You can grab the n-th element by unpacking the t-uple with value deconstruction, either by a let construct, a match construct or a function definition:
let ivuple = (5, 2, 1, 1)
let squared_sum_let =
let (a,b,c,d) = ivuple in
a*a + b*b + c*c + d*d
let squared_sum_match =
match ivuple with (a,b,c,d) -> a*a + b*b + c*c + d*d
let squared_sum_fun (a,b,c,d) =
a*a + b*b + c*c + d*d
The match-construct has here no virtue over the let-construct, it is just included for the sake of completeness.
Do not use t-uples, Don¹
There are only a few cases where using t-uples to represent a type is the right thing to do. Most of the times, we pick a t-uple because we are too lazy to define a type and we should interpret the problem of accessing the n-th field of a t-uple or iterating over the fields of a t-uple as a serious signal that it is time to switch to a proper type.
There are two natural replacements to t-uples: records and arrays.
When to use records
We can see a record as a t-uple whose entries are labelled; as such, they are definitely the most natural replacement to t-uples if we want to access them directly.
type ivuple = {
a: int;
b: int;
c: int;
d: int;
}
We then access directly the field a of a value x of type ivuple by writing x.a. Note that records are easily copied with modifications, as in let y = { x with d = 0 }. There is no natural way to iterate over the fields of a record, mostly because a record do not need to be homogeneous.
When to use arrays
A large² homogeneous collection of values is adequately represented by an array, which allows direct access, iterating and folding. A possible inconvenience is that the size of an array is not part of its type, but for arrays of fixed size, this is easily circumvented by introducing a private type — or even an abstract type. I described an example of this technique in my answer to the question “OCaml compiler check for vector lengths”.
Note on float boxing
When using floats in t-uples, in records containing only floats and in arrays, these are unboxed. We should therefore not notice any performance modification when changing from one type to the other in our numeric computations.
¹ See the TeXbook.
² Large starts near 4.
Since the length of OCaml tuples is part of the type and hence known (and fixed) at compile time, you get the n-th item by straightforward pattern matching on the tuple. For the same reason, the problem of extracting the n-th element of an "arbitrary-length tuple" cannot occur in practice - such a "tuple" cannot be expressed in OCaml's type system.
You might still not want to write out a pattern every time you need to project a tuple, and nothing prevents you from generating the functions get_1_1...get_i_j... that extract the i-th element from a j-tuple for any possible combination of i and j occuring in your code, e.g.
let get_1_1 (a) = a
let get_1_2 (a,_) = a
let get_2_2 (_,a) = a
let get_1_3 (a,_,_) = a
let get_2_3 (_,a,_) = a
...
Not necessarily pretty, but possible.
Note: Previously I had claimed that OCaml tuples can have at most length 255 and you can simply generate all possible tuple projections once and for all. As #Virgile pointed out in the comments, this is incorrect - tuples can be huge. This means that it is impractical to generate all possible tuple projection functions upfront, hence the restriction "occurring in your code" above.
It's not possible to write such a function in full generality in OCaml. One way to see this is to think about what type the function would have. There are two problems. First, each size of tuple is a different type. So you can't write a function that accesses elements of tuples of different sizes. The second problem is that different elements of a tuple can have different types. Lists don't have either of these problems, which is why you can have List.nth.
If you're willing to work with a fixed size tuple whose elements are all the same type, you can write a function as shown by #user2361830.
Update
If you really have collections of values of the same type that you want to access by index, you should probably be using an array.
here is a function wich return you the string of the ocaml function you need to do that ;) very helpful I use it frequently.
let tup len n =
if n>=0 && n<len then
let rec rep str nn = match nn<1 with
|true ->""
|_->str ^ (rep str (nn-1))in
let txt1 ="let t"^(string_of_int len)^"_"^(string_of_int n)^" tup = match tup with |" ^ (rep "_," n) ^ "a" and
txt2 =","^(rep "_," (len-n-2)) and
txt3 ="->a" in
if n = len-1 then
print_string (txt1^txt3)
else
print_string (txt1^txt2^"_"^txt3)
else raise (Failure "Error") ;;
For example:
tup 8 6;;
return:
let t8_6 tup = match tup with |_,_,_,_,_,_,a,_->a
and of course:
val t8_6 : 'a * 'b * 'c * 'd * 'e * 'f * 'g * 'h -> 'g = <fun>
I have the following code which I intend to create a Map with self defined types variable and location. I understand that the key type should be ordered (some comparator function). How shall I add these rules to make this work? Also, I find the code ugly here. Do I really need the ;; at the end of a type and module?
type variable = string;;
type location = int;;
module LocationMap = Map.Make(variable);;
module EnvironmentMap = Map.Make(location);;
EDIT
This is the rest of my code:
type variable = Variable of string
type location = Location of int
module LocationMap = Map.Make(struct type t = variable let compare = compare end)
module EnvironmentMap = Map.Make(struct type t = variable let compare = compare end)
(*file read function*)
let read_file filename =
let lines = ref [] in
let chan = open_in filename in
try
while true do
lines := input_line chan :: !lines
done;
!lines
with End_of_file ->
close_in chan;
List.rev !lines
in
(*get the inputs*)
let inputs = read_file Sys.argv.(1) in
for i = 0 to List.length inputs - 1 do
Printf.printf "%s\n" (List.nth inputs i)
done;
This has a syntax error. I am not sure why.
EDIT2
I make this work with the following edit:
type variable = Variable of string
type location = Location of int
module LocationMap = Map.Make(struct type t = variable let compare = compare end)
module EnvironmentMap = Map.Make(struct type t = variable let compare = compare end)
(*file read function*)
let read_file filename =
let lines = ref [] in
let chan = open_in filename in
try
while true do
lines := input_line chan :: !lines
done;
!lines
with End_of_file ->
close_in chan;
List.rev !lines
(*get the inputs*)
let () =
let inputs = read_file Sys.argv.(1) in
for i = 0 to List.length inputs - 1 do
Printf.printf "%s\n" (List.nth inputs i)
done;
Sorry for the long list of questions, what does let () = do here? Is it true that when I define a function with let, I do not need in?
When applying the Map.Make functor, you need to supply a struct containing your type and a compare function:
module LocationMap =
Map.Make(struct type t = variable let compare = compare end)
module EnvironmentMap =
Map.Make(struct type t = location let compare = compare end)
You never need to use ;; in compiled code. It's only required when using the toplevel, to tell it when it should evaluate what you've typed in so far.
Some people do use ;; in compiled code, but you never need to do this and I personally never do. There is always a way to get the same effect without using ;;.
Update
The let compare = compare binds the pre-existing OCaml function compare (the infamous polymorphic comparison function) to the name compare inside the struct. So, it creates a Map that uses polymorphic compare to do its comparisons. This is often what you want.
I created a file containing your definitions (without ;;) and the above code, then compiled it with ocamlc -c. There were no syntax errors. I'm positive you don't need to use ;;, as I've written many many thousands of lines of code without it.
Note that I'm not saying that if you remove ;; from syntactically correct OCaml code, the result is always syntactically correct. There are a few idioms that only work when you use ;;. I personally just avoid those idioms.
Update 2
A let at top level of a module is special, and doesn't have an in. It defines a global value of the module. OCaml treats every source file as a module (for free, as I like to say), with a name that's the same as the source file name (capitalized).
You can actually have any pattern in let pattern = expression. So let () = ... is completely normal. It just says that the expression has unit type (hence the pattern matches).
For a homework assignment, we've been instructed to complete a task without introducing any "side-effects". I've looked up "side-effects" on Wikipedia, and though I get that in theory it means "modifies a state or has an observable interaction with calling functions", I'm having trouble figuring out specifics.
For example, would creating a value that holds a non-compile time result be introducing side effects?
Say I had (might not be syntactically perfect):
val myList = (someFunction x y);;
if List.exists ((=) 7) myList then true else false;;
Would this introduce side-effects? I guess maybe I'm confused on what "modifies a state" means in the definition of side-effects.
No; a side-effect refers to e.g. mutating a ref cell with the assignment operator :=, or other things where the value referred to by a name changes over time. In this case, myList is an immutable value that never changes during the program, thus it is effect-free.
See also
http://en.wikipedia.org/wiki/Referential_transparency_(computer_science)
A good way to think about it is "have I changed anything which any later code (including running this same function again later) could ever possibly see other than the value I'm returning?" If so, that's a side effect. If not, then you can know that there isn't one.
So, something like:
let inc_nosf v = v+1
has no side effects because it just returns a new value which is one more than an integer v. So if you run the following code in the ocaml toplevel, you get the corresponding results:
# let x = 5;;
val x : int = 5
# inc_nosf x;;
- : int = 6
# x;;
- : int = 5
As you can see, the value of x didn't change. So, since we didn't save the return value, then nothing really got incremented. Our function itself only modifies the return value, not x itself. So to save it into x, we'd have to do:
# let x = inc_nosf x;;
val x : int = 6
# x;;
- : int = 6
Since the inc_nosf function has no side effects (that is, it only communicates with the outside world using its return value, not by making any other changes).
But something like:
let inc_sf r = r := !r+1
has side effects because it changes the value stored in the reference represented by r. So if you run similar code in the top level, you get this, instead:
# let y = ref 5;;
val y : int ref = {contents = 5}
# inc_sf y;;
- : unit = ()
# y;;
- : int ref = {contents = 6}
So, in this case, even though we still don't save the return value, it got incremented anyway. That means there must have been changes to something other than the return value. In this case, that change was the assignment using := which changed the stored value of the ref.
As a good rule of thumb, in Ocaml, if you avoid using refs, records, classes, strings, arrays, and hash tables, then you will avoid any risk of side effects. Although you can safely use string literals as long as you avoid modifying the string in place using functions like String.set or String.fill. Basically, any function which can modify a data type in place will cause a side effect.
[<ReflectedDefinition>]
let rec x = (fun() -> x + "abc") ()
The sample code with the recursive value above produces the following F# compiler error:
error FS0432: [<ReflectedDefinition>] terms cannot contain uses of the prefix splice operator '%'
I can't see any slicing operator usage in the code above, looks like a bug... :)
Looks like this is the problem with the quotation via ReflectedDefinitionAttribute only, normal quotation works well:
let quotation =
<# let rec x = (fun() -> x + "abc") () in x #>
produces expected result with the hidden Lazy.create and Lazy.force usages:
val quotation : Quotations.Expr<string> =
LetRecursive
([(x, Lambda (unitVar,
Application
(Lambda (unitVar0,
Call (None,
String op_Addition[String,String,String](String, String),
[Call (None,
String Force[String](Lazy`1[System.String]), // `
[x]), Value ("abc")])),
Value (<null>)))),
(x, Call (None, Lazy`1[String] Create[String](FSharpFunc`2[Unit,String]), [x])),
(x, Call (None, String Force[String](Lazy`1[String]), [x]))], x) // `
So the question is: is this an F# compiler bug or not?
I'd think that this may be caused by the treatment of recursive values in F#. As a workaround, you can turn the recursive reference into a parameter:
[<ReflectedDefinition>]
let foo x = (fun() -> x + "abc") ()
// To construct the recursive value, you'd write:
let rec x = foo x
The last line is of course invalid (just like your original code), because you're creating an immediate recursive reference, but it should give you the idea - in reality, you'd probably enclose x in a lambda function.
EDIT Originally, I thought that the problem may be as below, but I'm not sure now (see comments).
It looks more like a (probably known) limitation to me than an unexpected bug. There is an important difference between the two versions of the code you wrote - in the first case, you're binding a public value (visible to .NET) named x while in the second case, x is just a symbol used only in the quotation.
The quotation that would have to be stored in the meta-data of the assembly would look like this:
let rec x = <# (fun() -> %x + "abc") () #>
The body is quoted, but x is not a quoted symbol, so it needs to be spliced into the quotation (that is, it will be evaluated and the result will be used in its place). Note that this code will fail, because you're declaring a recursive value with immediate reference - x needs to be evaluated as part of its definition, so this won't work.
However, I think that % cannot appear in ReflectedDefinition quotations (that is, you cannot store the above in meta-data), because it involves some runtime aspects - you'd need to evaluate x when loading the meta-data.