I have written a member function, to check if an element exists in a list.
member(X, [X|_]).
member(X, [_|Y]) :- member(X, Y).
I load the .pl file into SWI prolog, and I get no warnings or errors pertaining to member. I test the member function using...
member(A, [1,2,3,4]).
This, obviously should return false. Instead I get
A = 1
Then when I try to enter new commands, the ide just shows me the key I typed, and says unknown action "what ever key i pressed"
I think my member function is sound, as it matches one written by my professor.
Any ideas?
You have called predicate member/2 in the i/o mode(o, i) e.i. first argument free (unbound) variable and the second is bound variable.
In that mode (despite its name) member/2 actually doesn't test element against list membership but simply enumerates list elements one-by-one and, finally, when no elements are remained, it fails (predicates fail in terms of logic programming, not return false ).
This predicate often used in so called failure driven control technique, e.g.:
trivial example when enumerated list elements are simply written
?- forall( member(X, [1,2,3,4]), write( X ) ).
Note, in this case will be no failure after list is exhausted.
To ensure this, the last clause is added, such that is meaning "the predicate successed at all".
p( List, Action ) :- generator( Variant, List ), side_effect_action( Variant ), fail.
p( _ , _ ).
generator( X, L ) :- member( X, L ).
Related
The target is to have the key j doing a possibly complex task and moving to the next line (the latter action performed just like the original function of the j key).
My initial attempt was to map j key this way:
nn j :<C-U>execute "call MyFun(" . v:count . ")"<CR>
(as you can see I intend to make j's behavior depend on the count which is prepended to it)
and to define the function MyFun appropriately:
fu! MyFun(count)
" do more stuff based on a:count
normal j
endf
which is faulty, as hitting j now results in the error E169: Command too recursive, since the non-recursivity of nnoremap, as long as my deduction is correct, applies to the "literal" content of the {rhs} of the mapping, and not to whatever is "inside" it (in other words the function body makes use of the meaning of j at the moment it is called, thus causing the infinte recursion).
Therefore I tried the following
nn , j
nn j :<C-U>execute "call MyFun(" . v:count . ")"<CR>
fu! MyFun(count)
" do more stuff based on a:count
normal ,
endf
However this means that I waste the key ,. I know I can avoid the waste of that mapping doing
nn <Plug>Nobody j
but then I wouldn't know how to use <Plug>Nobody (my understanding is indeed that its use is only in the {rhs} of another, non-nore mapping).
My initial attempt was to map j key this way
Using execute here is redundant. It's enough to do:
nnoremap j :<C-U>call MyFun(v:count)<CR>
now results in the error E169: Command too recursive
That's because of normal. To suppress remapping you must use "bang"-form: normal! j. Please, refere
to documentation for :normal, whose second paragraph describes exactly your use case:
If the [!] is given, mappings will not be used. Without it, when this
command is called from a non-remappable mapping (:noremap), the
argument can be mapped anyway.
Besides, note that j normally supports count, so 2j is expected to move two lines down. So you, probably, should do execute 'normal!' a:count . 'j' instead.
Suppose I have to update a list with each call to a function, such that, the previous element of the list are preserved.
Here is my attempt:
local
val all_list = [];
in
fun insert (x:int) : string = int2string (list_len( ((all_list#[x])) ) )
end;
The problem is that each time I call to insert, I get the output "1", which indicates that the list is initiated to [] again.
However I was expecting output of "1" for the first call to insert, and "2" for the second call,...etc.
I am not able to come with a workaround. How should it be done?
You need to use side-effects.
Most of the time functional programmers prefer to use pure functions, which don't have side effects. Your implementation is a pure function, so it will always return the same value for the same input (and in your case it returns the same value for any input).
You can deal with that by using a reference.
A crash course on references in Standard ML:
use ref to create a new reference, ref has type 'a -> 'a ref, so it packs an arbitrary value into a reference cell, which you can modify later;
! is for unpacking a reference: (!) : 'a ref -> 'a, in most imperative languages this operation is implicit, but not in SML or OCaml;
(:=) : 'a ref * 'a -> unit is an infix operator used for modifying references, here is how you increment the contents of an integer reference: r := !r + 1.
The above gives us the following code (I prepend xs onto the list, instead of appending them):
local
val rxs = ref []
in
fun insert (x:int) : string =
(rxs := x :: !rxs;
Int.toString (length (!rxs)))
end
Values are immutable in SML. insert is defined in a context in which the value of all_list is [] and nothing about your code changes that value.
all_list#[x]
doesn't mutate the value all_list -- it returns a brand new list, one which your code promptly discards (after taking its length).
Using reference types (one of SML's impure features) it is possible to do what you seem to be trying to do, but the resulting code wouldn't be idiomatic SML. It would break referential transparency (the desirable feature of functional programming languages where a function called with identical inputs yields identical outputs).
I'm learning Erlang from the very basic and have a problem with a tail recursive function. I want my function to receive a list and return a new list where element = element + 1. For example, if I send [1,2,3,4,5] as an argument, it must return [2,3,4,5,6]. The problem is that when I send that exact arguments, it returns [[[[[[]|2]|3]|4]|5]|6].
My code is this:
-module(test).
-export([test/0]).
test()->
List = [1,2,3,4,5],
sum_list_2(List).
sum_list_2(List)->
sum_list_2(List,[]).
sum_list_2([Head|Tail], Result)->
sum_list_2(Tail,[Result|Head +1]);
sum_list_2([], Result)->
Result.
However, if I change my function to this:
sum_list_2([Head|Tail], Result)->
sum_list_2(Tail,[Head +1|Result]);
sum_list_2([], Result)->
Result.
It outputs [6,5,4,3,2] which is OK. Why the function doesn't work the other way around([Result|Head+1] outputing [2,3,4,5,6])?
PS: I know this particular problem is solved with list comprehensions, but I want to do it with recursion.
For this kind of manipulation you should use list comprehension:
1> L = [1,2,3,4,5,6].
[1,2,3,4,5,6]
2> [X+1 || X <- L].
[2,3,4,5,6,7]
it is the fastest and most idiomatic way to do it.
A remark on your fist version: [Result|Head +1] builds an improper list. the construction is always [Head|Tail] where Tail is a list. You could use Result ++ [Head+1] but this would perform a copy of the Result list at each recursive call.
You can also look at the code of lists:map/2 which is not tail recursive, but it seems that actual optimization of the compiler work well in this case:
inc([H|T]) -> [H+1|inc(T)];
inc([]) -> [].
[edit]
The internal and hidden representation of a list looks like a chained list. Each element contains a term and a reference to the tail. So adding an element on top of the head does not need to modify the existing list, but adding something at the end needs to mutate the last element (the reference to the empty list is replaced by a reference to the new sublist). As variables are not mutable, it needs to make a modified copy of the last element which in turn needs to mutate the previous element of the list and so on. As far as I know, the optimizations of the compiler do not make the decision to mutate variable (deduction from the the documentation).
The function that produces the result in reverse order is a natural consequence of you adding the newly incremented element to the front of the Result list. This isn't uncommon, and the recommended "fix" is to simply list:reverse/1 the output before returning it.
Whilst in this case you could simply use the ++ operator instead of the [H|T] "cons" operator to join your results the other way around, giving you the desired output in the correct order:
sum_list_2([Head|Tail], Result)->
sum_list_2(Tail, Result ++ [Head + 1]);
doing so isn't recommended because the ++ operator always copies it's (increasingly large) left hand operand, causing the algorithm to operate in O(n^2) time instead of the [Head + 1 | Tail] version's O(n) time.
I am beginner of Prolog.
what I have is a function traverse a list and return true when it satisfies the condition.
for example, check_version checks if the package version met the condition(eg. the version satisfies the condition such as greater than or less than the specific version) and check_all checks takes a list of versions and conditions to check one by one.
package('python', '2.6.5').
package('python', '2.5.4').
package('python', '1.5.2').
package('python', '3.1.0').
check_version(Pac, Ver, Cmp, V):-
package(Pac, V),
cmp_version(V, Ver, Cmp).
check_all( Pac, [], [], V):-
package(Pac, V).
check_all(Pac, [Ver], [Cmp], V):-
check_version(Pac, Ver, Cmp, V).
check_all(Pac, [Ver|VerS], [Cmp|CmpS], V):-
check_version(Pac, Ver, Cmp, V),
check_all(Pac, VerS, CmpS, V).
The problem I have is when try to find other solutions, it gives me duplicate solution.
I get:
check_all('python', ['3.0','2.4'], [lt,ge], V).
V = '2.6.5' ;
V = '2.6.5' ;
V = '2.5.4' ;
V = '2.5.4' .
expected:
check_all('python', ['3.0','2.4'], [lt,ge], V).
V = '2.6.5' ;
V = '2.5.4' .
I used trace to track it, and the problem I found, when it try to find another solution it back tracks and will return fail until find the right solution. Like the example above, apparently, it will return true for V='2.6.5' at first and take that to back track and run the functions, and we expect it returns false and then when it reach the beginning it run package('python', V) and V will take another value.
...
Exit: (7) check_all(python, ['3.0', '2.4'], [lt, ge], '2.6.5') ? creep
V = '2.6.5'
...
Fail: (9) check_version(python, '2.4', ge, '2.6.5') ? creep
Redo: (8) check_all(python, ['2.4'], [ge], '2.6.5') ? creep
Call: (9) check_version(python, '2.4', ge, '2.6.5') ? creep
Call: (10) package(python, '2.6.5') ? creep
Exit: (10) package(python, '2.6.5') ? creep
when back tracking, in check_all, it fails at check_all as we expected, but it returns true when it backtracks check_version and run package(python, '2.6.5') as V=2.6.5 a new value. so it return true again when V=2.6.5. is there any way to solve this problem?
To localize your problem, first reduce the size of your input. A single element suffices:
?- check_all('python', ['3.0'], [lt], V).
Now, which rules apply for a single element?
Both rules apply! So remove the more specialized one.
There is also another way how to localize such a problem. Simply compare the rules to each other and try to figure out a case where they both apply. The last rule applies for VerS = [] when also the first applies.
Applying a predicate to each element of a list is best done by a predicate that has the list as its first argument. Without going into detail, this makes the predicate succeed when the iteration is complete, if the argument is a list and not a variable (i.e. when it is an input argument). You should have two clauses: one to deal with the empty list and one for the general case:
foo([]). % succeed
foo([X|Xs]) :-
/* apply a predicate to X */
foo(Xs). % apply predicate to the rest of the list
An important thing here is that you don't need a third clause that deals with lists with one element only, since a list with one element is actually a list with an element and an empty list as its tail:
?- [a] == [a|[]].
true.
?- [a] = [a|[]].
true.
Another important thing is that there is nothing you should be doing in the base case, empty list (at least for your example).
To the problem now: your inputs are
the package name
two lists holding pairs of arguments to a predicate you have defined elsewhere (cmp_version/3). This is your list of conditions.
Implementation:
Known packages are available as facts: they can be enumerated by backtracking.
Conditions are an input arguments, provided as a list: you need to apply the condition to each element of the list(s).
The predicate:
check_all([], [], _, _).
check_all([V|Vs], [C|Cs], Name, Version) :-
package(Name, V), % enumerate all known packages by backtracking
cmp_version(Version, V, Cmp), % condition
check_all(Vs, Cs, Name, Version). % apply condition to the rest of the list(s)
You should read the documentation of maplist. You can express the query for example as:
?- maplist(check_version(python), ['3.0', '2.4'], [lt, ge], Versions).
where you have defined a predicate check_version/4 that looks something like:
check_version(Name, V, Cmp, Version) :-
package(Name, Version),
cmp_version(Version, V, Cmp).
As a side note, maplist will reorder its arguments to make it behave like the explicit iteration above.
EDIT
Naming issues, after #mat's comments: one very useful naming convention is to use a name that has descriptive one-word names for the arguments, delimited by underscores. For example, package/2 becomes package_version/2 since its first argument is the package and the second one the version.
I've read through previous topics on closures on stackflow and other sources and one thing is still confusing me. From what I've been able to piece together technically a closure is simply the set of data containing the code of a function and the value of bound variables in that function.
In other words technically the following C function should be a closure from my understanding:
int count()
{
static int x = 0;
return x++;
}
Yet everything I read seems to imply closures must somehow involve passing functions as first class objects. In addition it usually seems to be implied that closures are not part of procedural programming. Is this a case of a solution being overly associated with the problem it solves or am I misunderstanding the exact definition?
No, that's not a closure. Your example is simply a function that returns the result of incrementing a static variable.
Here's how a closure would work:
function makeCounter( int x )
{
return int counter() {
return x++;
}
}
c = makeCounter( 3 );
printf( "%d" c() ); => 4
printf( "%d" c() ); => 5
d = makeCounter( 0 );
printf( "%d" d() ); => 1
printf( "%d" c() ); => 6
In other words, different invocations of makeCounter() produce different functions with their own binding of variables in their lexical environment that they have "closed over".
Edit: I think examples like this make closures easier to understand than definitions, but if you want a definition I'd say, "A closure is a combination of a function and an environment. The environment contains the variables that are defined in the function as well as those that are visible to the function when it was created. These variables must remain available to the function as long as the function exists."
For the exact definition, I suggest looking at its Wikipedia entry. It's especially good. I just want to clarify it with an example.
Assume this C# code snippet (that's supposed to perform an AND search in a list):
List<string> list = new List<string> { "hello world", "goodbye world" };
IEnumerable<string> filteredList = list;
var keywords = new [] { "hello", "world" };
foreach (var keyword in keywords)
filteredList = filteredList.Where(item => item.Contains(keyword));
foreach (var s in filteredList) // closure is called here
Console.WriteLine(s);
It's a common pitfall in C# to do something like that. If you look at the lambda expression inside Where, you'll see that it defines a function that it's behavior depends on the value of a variable at its definition site. It's like passing a variable itself to the function, rather than the value of that variable. Effectively, when this closure is called, it retrieves the value of keyword variable at that time. The result of this sample is very interesting. It prints out both "hello world" and "goodbye world", which is not what we wanted. What happened? As I said above, the function we declared with the lambda expression is a closure over keyword variable so this is what happens:
filteredList = filteredList.Where(item => item.Contains(keyword))
.Where(item => item.Contains(keyword));
and at the time of closure execution, keyword has the value "world," so we're basically filtering the list a couple times with the same keyword. The solution is:
foreach (var keyword in keywords) {
var temporaryVariable = keyword;
filteredList = filteredList.Where(item => item.Contains(temporaryVariable));
}
Since temporaryVariable is scoped to the body of the foreach loop, in every iteration, it is a different variable. In effect, each closure will bind to a distinct variable (those are different instances of temporaryVariable at each iteration). This time, it'll give the correct results ("hello world"):
filteredList = filteredList.Where(item => item.Contains(temporaryVariable_1))
.Where(item => item.Contains(temporaryVariable_2));
in which temporaryVariable_1 has the value of "hello" and temporaryVariable_2 has the value "world" at the time of closure execution.
Note that the closures have caused an extension to the lifetime of variables (their life were supposed to end after each iteration of the loop). This is also an important side effect of closures.
From what I understand a closure also has to have access to the variables in the calling context. Closures are usually associated with functional programming. Languages can have elements from different types of programming perspectives, functional, procedural, imperative, declarative, etc. They get their name from being closed over a specified context. They may also have lexical binding in that they can reference the specified context with the same names that are used in that context. Your example has no reference to any other context but a global static one.
From Wikipedia
A closure closes over the free variables (variables which are not local variables)
A closure is an implementation technique for representing procedures/functions with local state. One way to implement closures is described in SICP. I will present the gist of it, anyway.
All expressions, including functions are evaluated in an environement, An environment is a sequence of frames. A frame maps variable names to values. Each frame also has a
pointer to it's enclosing environment. A function is evaluated in a new environment with a frame containing bindings for it's arguments. Now let us look at the following interesting scenario. Imagine that we have a function called accumulator, which when evaluated, will return another function:
// This is some C like language that has first class functions and closures.
function accumulator(counter) {
return (function() { return ++counter; });
}
What will happen when we evaluate the following line?
accum1 = accumulator(0);
First a new environment is created and an integer object (for counter) is bound to 0 in it's first frame. The returned value, which is a new function, is bound in the global environment. Usually the new environment will be garbage collected once the function
evaluation is over. Here that will not happen. accum1 is holding a reference to it, as it needs access to the variable counter. When accum1 is called, it will increment the value of counter in the referenced environment. Now we can call accum1 a function with local state or a closure.
I have described a few practical uses of closures at my blog
http://vijaymathew.wordpress.com. (See the posts "Dangerous designs" and "On Message-Passing").
There's a lot of answers already, but I'll add another one anyone...
Closures aren't unique to functional languages. They occur in Pascal (and family), for instance, which has nested procedures. Standard C doesn't have them (yet), but IIRC there is a GCC extension.
The basic issue is that a nested procedure may refer to variables defined in it's parent. Furthermore, the parent may return a reference to the nested procedure to its caller.
The nested procedure still refers to variables that were local to the parent - specifically to the values those variables had when the line making the function-reference was executed - even though those variables no longer exist as the parent has exited.
The issue even occurs if the procedure is never returned from the parent - different references to the nested procedure constructed at different times may be using different past values of the same variables.
The resolution to this is that when the nested function is referenced, it is packaged up in a "closure" containing the variable values it needs for later.
A Python lambda is a simple functional-style example...
def parent () :
a = "hello"
return (lamda : a)
funcref = parent ()
print funcref ()
My Pythons a bit rusty, but I think that's right. The point is that the nested function (the lambda) is still referring to the value of the local variable a even though parent has exited when it is called. The function needs somewhere to preserve that value until it's needed, and that place is called a closure.
A closure is a bit like an implicit set of parameters.
Great question! Given that one of the OOP principles of OOP is that objects has behavior as well as data, closures are a special type of object because their most important purpose is their behavior. That said, what do I mean when I talk about their "behavior?"
(A lot of this is drawn from "Groovy in Action" by Dierk Konig, which is an awesome book)
On the simplest level a close is really just some code that's wrapped up to become an androgynous object/method. It's a method because it can take params and return a value, but it's also an object in that you can pass around a reference to it.
In the words of Dierk, imagine an envelope that has a piece of paper inside. A typical object would have variables and their values written on this paper, but a closure would have a list of instructions instead. Let's say the letter says to "Give this envelope and letter to your friends."
In Groovy: Closure envelope = { person -> new Letter(person).send() }
addressBookOfFriends.each (envelope)
The closure object here is the value of the envelope variable and it's use is that it's a param to the each method.
Some details:
Scope: The scope of a closure is the data and members that can be accessed within it.
Returning from a closure: Closures often use a callback mechanism to execute and return from itself.
Arguments: If the closure needs to take only 1 param, Groovy and other langs provide a default name: "it", to make coding quicker.
So for example in our previous example:
addressBookOfFriends.each (envelope)
is the same as:
addressBookOfFriends.each { new Letter(it).send() }
Hope this is what you're looking for!
An object is state plus function.
A closure, is function plus state.
function f is a closure when it closes over (captured) x
I think Peter Eddy has it right, but the example could be made more interesting. You could define two functions which close over a local variable, increment & decrement. The counter would be shared between that pair of functions, and unique to them. If you define a new pair of increment/decrement functions, they would be sharing a different counter.
Also, you don't need to pass in that initial value of x, you could let it default to zero inside the function block. That would make it clearer that it's using a value which you no longer have normal access to otherwise.