I have two questions concerning OCaml.
Firstly, what does the == means when defining a type.
For example you can see at the end of this page the following code:
type compteur == int;;
Then what is the difference with:
type compteur = int;;
Moreover I have an other question concerning pattern matching.
How to say that you want to return nothing on a case.
For example let's say I have a function f that returns a boolean:
let rec f v = function
| t when t<v -> true
| t when t > v -> f (t-1)
| t when t = v -> (* here a code to say that you do nothing, and wait for the other recursive call *)
type compteur == int is a syntax error. The only valid way to define a type alias is with =, not ==. It's just a typo on the page you linked.
How to say that you want to return nothing on a case.
The only way to return nothing from a function would be to exit the program, raise an exception or loop (or recur) infinitely. Otherwise a function always returns a value.
here a code to say that you do nothing, and wait for the other recursive call
What other recursive call? In the case that t = v only the code for that case will run. There is no other code to wait on.
I've read a number of articles on the difference between assignment and binding, but it hasn't clicked yet (specifically in the context of an imperative language vs one without mutation).
I asked in IRC, and someone mentioned these 2 examples illustrate the difference, but then I had to go and I didn't see the full explanation.
Can someone please explain how/why this works in a detailed way, to help illustrate the difference?
Ruby
x = 1; f = lambda { x }; x = 2; f.call
#=> 2
Elixir
x = 1; f = fn -> x end; x = 2; f.()
#=> 1
I've heard this explanation before and it seems pretty good:
You can think of binding as a label on a suitcase, and assignment as a
suitcase.
In other languages, where you have assignment, it is more like putting a value in a suitcase. You actually change value that is in the suitcase and put in a different value.
If you have a suitcase with a value in it, in Elixir, you put a label on it. You can change the label, but the value in the suitcase is still the same.
So, for example with:
iex(1)> x = 1
iex(2)> f = fn -> x end
iex(3)> x = 2
iex(4)> f.()
1
You have a suitcase with 1 in it and you label it x.
Then you say, "Here, Mr. Function, I want you to tell me what is in this suitcase when I call you."
Then, you take the label off of the suitcase with 1 in it and put it on another suitcase with 2 in it.
Then you say "Hey, Mr. Function, what is in that suitcase?"
He will say "1", because the suitcase hasn't changed. Although, you have taken your label off of it and put it on a different suitcase.
After a while, I came up with the answer that is probably the best explanation of the difference between “binding” and “assignment”; it has nothing in common with what I have written in another answer, hence it’s posted as a separate answer.
In any functional language, where everything is immutable, there is no meaningful difference between terms “binding” and “assignment.” One might call it either way; the common pattern is to use the word “binding,“ explicitly denoting that it’s a value bound to a variable. In Erlang, for instance, one can not rebound a variable. In Elixir this is possible (why, for God’s sake, José, what for?)
Consider the following example in Elixir:
iex> x = 1
iex> 1 = x
The above is perfectly valid Elixir code. It is evident, one cannot assign anything to one. It is neither assignment nor binding. It is matching. That is how = is treated in Elixir (and in Erlang): a = b fails if both are bound to different values; it returns RHO if they match; it binds LHO to RHO if LHO is not bound yet.
In Ruby it differs. There is a significant difference between assignment (copying the content,) and binding (making a reference.)
Elixir vs Ruby might not be the best contrast for this. In Elixir, we can readily "re-assign" the value of a previously assigned named variable. The two anonymous-function examples you provided demonstrate the difference in how the two languages assign local variables in them. In Ruby, the variable, meaning the memory reference, is assigned, which is why when we change it, the anonymous function returns the current value stored in that memory-reference. While in Elixir, the value of the variable at the time the anonymous function is defined (rather than the memory reference) is copied and stored as the local variable.
In Erlang, Elixir's "parent" language, however, variables as a rule are "bound." Once you've declared the value for the variable named X, you are not allowed to alter it for the remainder of the program and any needed alterations would need to be stored in new named variables. (There is a way to reassign a named variable in Erlang but it is not the custom.)
Binding refers to particular concept used in expression-based languages that may seem foreign if you're used to statement-based languages. I'll use an ML-style example to demonstrate:
let x = 3 in
let y = 5 in
x + y
val it : int = 8
The let... in syntax used here demonstrates that the binding let x = 3 is scoped only to the expression following the in. Likewise, the binding let y = 5 is only scoped to the expression x + y, such that, if we consider another example:
let x = 3 in
let f () =
x + 5
let x = 4 in
f()
val it : int = 8
The result is still 8, even though we have the binding let x = 4 above the call to f(). This is because f itself was bound in the scope of the binding let x = 3.
Assignment in statement-based languages is different, because the variables being assigned are not scoped to a particular expression, they are effectively 'global' for whatever block of code they're in, so reassigning the value of a variable changes the result of an evaluation that uses the same variable.
The easiest way to understand the difference, would be to compare the AST that is used by the language interpreter/compiler to produce machine-/byte-code.
Let’s start with ruby. Ruby does not provide the AST viewer out of the box, so I will use RubyParser gem for that:
> require 'ruby_parser'
> RubyParser.new.parse("x = 1; f = -> {x}; x = 2; f.()").inspect
#=> "s(:block, s(:lasgn, :x, s(:lit, 1)),
# s(:lasgn, :f, s(:iter, s(:call, nil, :lambda), 0, s(:lvar, :x))),
# s(:lasgn, :x, s(:lit, 2)), s(:call, s(:lvar, :f), :call))"
The thing we are looking for is the latest node in the second line: there is x variable inside the proc. In other words, ruby expects the bound variable there, named x. At the time the proc is evaluated, x has a value of 2. Hence the the proc returns 2.
Let’s now check Elixir.
iex|1 ▶ quote do
...|1 ▶ x = 1
...|1 ▶ f = fn -> x end
...|1 ▶ x = 2
...|1 ▶ f.()
...|1 ▶ end
#⇒ {:__block__, [],
# [
# {:=, [], [{:x, [], Elixir}, 1]},
# {:=, [], [{:f, [], Elixir}, {:fn, [], [{:->, [], [[], {:x, [], Elixir}]}]}]},
# {:=, [], [{:x, [], Elixir}, 2]},
# {{:., [], [{:f, [], Elixir}]}, [], []}
# ]}
Last node in the second line is ours. It still contains x, but during a compilation stage this x will be evaluated to it’s currently assigned value. That said, fn -> not_x end will result in compilation error, while in ruby there could be literally anything inside a proc body, since it’ll be evaluated when called.
In other words, Ruby uses a current caller’s context to evaluate proc, while Elixir uses a closure. It grabs the context it encountered the function definition and uses it to resolve all the local variables.
def success?
return #fhosts.empty? and #khosts.empty? and #shosts.any?
end
When I run that instance method, I get an error:
/home/fandingo/code/management/lib/ht.rb:37: void value expression
return #fhosts.empty? and #khosts.empty? and #shosts.any?
I'm confused by what's happening since this works
def success?
#fhosts.empty? and #khosts.empty? and #shosts.any?
# This also works
# r = #fhosts.empty? and #khosts.empty? and #shosts.any?
# return r
end
I'm coming from a Python background, and I don't want anything to do with implicit returns. Programming has plenty of landmines as it is.
If we have an arbitrary expression, E, that consists of boolean operations and and or together, here are some operations we could perform:
if E -- works
E -- works
* v = E -- works
return E -- broken
Why doesn't the last case work?
Edit: Actually v = E doesn't work. Only
v = Ei
is evaluated. Ei+1...k are ignored.
This is likely due to the very weak binding of and which causes it to parse out differently than you expect:
return x and y
This actually means:
(return x) and y
Since you're returning immediately it doesn't have a chance to evaluate the remainder of the expression.
Your version without return is correct:
x and y
This doesn't have a binding issue and is more idiomatic Ruby. Remember you only need to put an explicit return if you're trying to force an exit before the last line of the method. Being opposed to implicit returns is going to make your code look heavily non-Ruby. They're one of the reasons Ruby is so clean and simple, and how things like a.map { |v| v * 2 } works.
The When in Rome principle applies here. If you want to write Python-style Ruby you're going to be going against the grain. It's like saying "I don't like how you say X in your spoken language, so I'll just ignore that and do it my way."
This should also work:
return x && y
The && method is very strongly bound so return is the last thing evaluated here.
Or if you really want to use and for whatever reason:
return (x and y)
Consider the following code snippet:
class Example
def my_attr=(value)
#_my_attr = value
#_my_attr * 3
end
end
I expect the expression Example.new.my_attr = 5 to return 15, but that turns out to be wrong. The original return value is always returned, even when I call the = method explicitly:
Example.new.my_attr = 5 # => 5
Example.new.my_attr=(5) # => 5
How and why does Ruby do this? Does Ruby treat methods that end in = specially, or is it some other mechanism? I guess this precludes chaining on return values of = methods, right? Is there a way to make Ruby behave differently, or is this just how it is?
Update: Credit to #jeffgran for this:
Example.new.send(:my_attr=, 5) # => 15
This is a workaround, but on another level even more perplexing, since that would mean send is clearly not always equivalent in behavior to calling a method directly.
This is how assignment works; the return value is ignored, and the result of an assignment expression is always the right-hand value. This is a fundamental feature of Ruby's grammar. left-hand side = right-hand side will always evaluate to right-hand side, regardless of whether left hand side is a variable (x), a method (object.x), a constant (X) or any expression.
Source: Programming Languages | Ruby
IPA Ruby Standardization WG Draft, 11.4.2.2.5, Single method assignments
Consider chaining of assignments, x = y = 3.
For this to work correctly, the result of y = 3 must be 3, regardless of the actual value returned by the y= method. x = y = 3 is meant to read as y = 3; x = 3, not as y = 3; x = y which is what would be implied if the return value from y= was treated as the result of y = 3.
Or consider all the other places assignment can be used. Sometimes, instead of this...
obj.x = getExpensiveThing()
if obj.x
...
... we write this ...
if obj.x = getExpensiveThing()
This couldn't work if the result of obj.x = ... could be any arbitrary thing, but we know it will work because the result of obj.x = y is always y.
Update
A comment on the question states:
Interesting, I wasn't aware of this scenario. It seems that method= returns whatever input is given...
No, it's an important distinction to make. This has nothing to do with the return value of method assignment, and it definitely does not "return whatever input is given", it returns whatever you tell it to return.
The whole point is that the return value is ignored by the grammar of the language; assignment doesn't evaluate to the return value of the attr= method, but the return value still exists as evidenced by the question itself: Example.new.send(:my_attr=, 5) # => 15. This works because it is not assignment. You're side-stepping that part of the Ruby language.
Update again
To be clear: x and y in my examples shouldn't be interpreted as literal Ruby variables, they are place holders for any valid left-hand side of an assignment. x or y could be any expression: a, obj.a, CONSTANT_A, Something::a, #instance_a, it's all the same. The value of assignment is always the right-hand side.
I'm trying to make a function that defines a vector that varies based on the function's input, and set! works great for this in Scheme. Is there a functional equivalent for this in OCaml?
I agree with sepp2k that you should expand your question, and give more detailed examples.
Maybe what you need are references.
As a rough approximation, you can see them as variables to which you can assign:
let a = ref 5;;
!a;; (* This evaluates to 5 *)
a := 42;;
!a;; (* This evaluates to 42 *)
Here is a more detailed explanation from http://caml.inria.fr/pub/docs/u3-ocaml/ocaml-core.html:
The language we have described so far is purely functional. That is, several evaluations of the same expression will always produce the same answer. This prevents, for instance, the implementation of a counter whose interface is a single function next : unit -> int that increments the counter and returns its new value. Repeated invocation of this function should return a sequence of consecutive integers — a different answer each time.
Indeed, the counter needs to memorize its state in some particular location, with read/write accesses, but before all, some information must be shared between two calls to next. The solution is to use mutable storage and interact with the store by so-called side effects.
In OCaml, the counter could be defined as follows:
let new_count =
let r = ref 0 in
let next () = r := !r+1; !r in
next;;
Another, maybe more concrete, example of mutable storage is a bank account. In OCaml, record fields can be declared mutable, so that new values can be assigned to them later. Hence, a bank account could be a two-field record, its number, and its balance, where the balance is mutable.
type account = { number : int; mutable balance : float }
let retrieve account requested =
let s = min account.balance requested in
account.balance <- account.balance -. s; s;;
In fact, in OCaml, references are not primitive: they are special cases of mutable records. For instance, one could define:
type 'a ref = { mutable content : 'a }
let ref x = { content = x }
let deref r = r.content
let assign r x = r.content <- x; x
set! in Scheme assigns to a variable. You cannot assign to a variable in OCaml, at all. (So "variables" are not really "variable".) So there is no equivalent.
But OCaml is not a pure functional language. It has mutable data structures. The following things can be assigned to:
Array elements
String elements
Mutable fields of records
Mutable fields of objects
In these situations, the <- syntax is used for assignment.
The ref type mentioned by #jrouquie is a simple, built-in mutable record type that acts as a mutable container of one thing. OCaml also provides ! and := operators for working with refs.