I am very new to F# programming. I am looking for an answer as to why the following recursive function will exit when the n=0 condition is reached. Does the "then 1" syntax have a special meaning that equates to EXIT?
let rec factorial n =
if n = 0
then 1
else n * factorial (n - 1)
Functional programming languages are expression oriented, as opposed to statement oriented. This means that everything is an expression that can be evaluated into a value.
The control flow structures, if and match, are also just expressions. The compiler checks that all branches of these expressions return the same type.
If you're familiar with the conditional ternary operator in other languages, F#'s expression if true then 1 else 0, is equivalent to true ? 1 : 0.
Furthermore, the function doesn't "exit" as such. It completes evaluation. Each instance of the function (there will be n + 1 instances) completes evaluation at the end of the if/then/else expression.
Related
On wikipedia it says that using call/cc you can implement the amb operator for nondeterministic choice, and my question is how would you implement the amb operator in a language in which the only support for continuations is to write in continuation passing style, like in erlang?
If you can encode the constraints for what constitutes a successful solution or choice as guards, list comprehensions can be used to generate solutions. For example, the list comprehension documentation shows an example of solving Pythagorean triples, which is a problem frequently solved using amb (see for example exercise 4.35 of SICP, 2nd edition). Here's the more efficient solution, pyth1/1, shown on the list comprehensions page:
pyth1(N) ->
[ {A,B,C} ||
A <- lists:seq(1,N-2),
B <- lists:seq(A+1,N-1),
C <- lists:seq(B+1,N),
A+B+C =< N,
A*A+B*B == C*C
].
One important aspect of amb is efficiently searching the solution space, which is done here by generating possible values for A, B, and C with lists:seq/2 and then constraining and testing those values with guards. Note that the page also shows a less efficient solution named pyth/1 where A, B, and C are all generated identically using lists:seq(1,N); that approach generates all permutations but is slower than pyth1/1 (for example, on my machine, pyth(50) is 5-6x slower than pyth1(50)).
If your constraints can't be expressed as guards, you can use pattern matching and try/catch to deal with failing solutions. For example, here's the same algorithm in pyth/1 rewritten as regular functions triples/1 and the recursive triples/5:
-module(pyth).
-export([triples/1]).
triples(N) ->
triples(1,1,1,N,[]).
triples(N,N,N,N,Acc) ->
lists:reverse(Acc);
triples(N,N,C,N,Acc) ->
triples(1,1,C+1,N,Acc);
triples(N,B,C,N,Acc) ->
triples(1,B+1,C,N,Acc);
triples(A,B,C,N,Acc) ->
NewAcc = try
true = A+B+C =< N,
true = A*A+B*B == C*C,
[{A,B,C}|Acc]
catch
error:{badmatch,false} ->
Acc
end,
triples(A+1,B,C,N,NewAcc).
We're using pattern matching for two purposes:
In the function heads, to control values of A, B and C with respect to N and to know when we're finished
In the body of the final clause of triples/5, to assert that conditions A+B+C =< N and A*A+B*B == C*C match true
If both conditions match true in the final clause of triples/5, we insert the solution into our accumulator list, but if either fails to match, we catch the badmatch error and keep the original accumulator value.
Calling triples/1 yields the same result as the list comprehension approaches used in pyth/1 and pyth1/1, but it's also half the speed of pyth/1. Even so, with this approach any constraint could be encoded as a normal function and tested for success within the try/catch expression.
If I have the following function:
let myFunc x y =
if y = 0 then 1
x
I get the error:
Program.fs(58,17): error FS0001: This expression was expected to have type
unit
but here has type
int
Why does the compiler expect 'unit' instead of int ?
It might worth adding that this is not just a property of if. F# is an expression-based language meaning that pretty much every piece of code (aside from type declarations and a few exceptions) is an expression that evaluates to some result. In fact, F# does not call if the if statement, but an if expression.
This means that you can use if in unexpected places. For example, this might be useful:
x/2 + (if x%2=0 then 0 else 1)
As already explained by Garry, if you omit else, then the expression still needs to return something - if the result was to be an int, then it would not really make sense (which number should the compiler pick?), so it requires that the result is of type unit, which is a special type representing "no result".
The unit type is also the result of all imperative functions (e.g. printf) or of all expressions that do not logically return any value (assignment or e.g. loop). This means that if you write:
if x > 0 then printfn "Big!"
... then the expression is well-typed, because printfn "Big!" has a return type unit and the implicitly added else branch also returns unit. You can create a value of type unit directly by hand (the type has exactly one value), so the above actually corresponds to:
if x > 0 then printfn "Big!" else ()
From the C# perspective, it makes more sense to read if .. then .. else as the conditional operator:
x/2 + (x%2 == 0 ? 0 : 1)
In F# when using an if statement when there is no else branch then it implicitly returns unit. If your then branch returns a type other than unit you must have an explicit else branch for it to work correctly. In your example you could write:
let myFunc x y = if y = 0 then 1 else x
MSDN - http://msdn.microsoft.com/en-us/library/dd233231.aspx
I am looking at the following F# line
for i = 0 to i=10 do
Console.WriteLine("Hello")
An I am wondering that isn't the above line a statement as opposed to an expression?
Shouldn't everything be an expression in F#?
As already said, every syntactical construct in F# is an expression. F# does not distinguish between statements and expressions (and so I'd say that the WikiPedia quote posted by Robert is a bit misleading - F# does not have statements).
Actually, the above is not fully true, because some constructs in F# computation expressions such as let! are not expressions, but we can ignore that.
What does that mean? In C#, the syntax of for and method calls is defined something like this:
statement := foreach(var v in <expression>) <statement>
| { <statement> ... <statement> }
| <expression>;
| (...)
expression := <expression>.<ident>(<expression>, ..., <expression>)
| <literal>
| <expression> + <expression>
| (...)
This is very simplified, but it should give you the idea - a statement is something that does not evaluate to a value. It can be foreach loop (other loops), a statement block (with multiple statements) or an expression with semicolon (where the result of the expression is void or is ignored). An expression is, for example, method call, primitive literal (string, int) or a binary operator.
This means that you cannot write certain things in C# - for example, the argument of method call cannot be a statement (because statements do not evaluate to a value!)
On the other hand, in F#, everything is an expression. This means there is just a single syntactic category:
expression := for v in <expression> do <expression>
| <expression>; <expression>
| <expression>.<ident>(<expression>, ..., <expression>)
| <literal>
| <expression> + <expression>
| (...)
This means that in F# all syntactic constructs are expressions, including for and other loops. The body of for is also an expression, but it would not make sense if the expression evaluated to some value (i.e. 42), so the types require that the result of the body is unit (which does not carry any information). Similarly, the first expression in sequencing (<expr>; <expr>) should return unit - the result of sequencing is the result of the second expression.
This makes the language simpler and more uniform, but you can write some odd things:
let x = (for i in 0 .. 10 do printfn "%d" i); 42
This will print numbers from 0 to 10 and then define a value x to be 42. The assignment is a sequencing of expressions (<expr>; <expr>) where the first one is for loop (that has a type unit, because it does not evaluate to anything) and the second one is 42, which evaluates to 42.
Every statement in F#, including if statements and loops, is a
composable expression with a definite return type.
Functions and expressions that do not return any value have a return
type of unit.
http://en.wikipedia.org/wiki/F_Sharp_(programming_language)
In languages like F# statements are just expressions that return the value () of type unit. As the unit type has only one value it conveys no information so returning the value of type unit is saying "if I'm doing anything then it is by way of a side effect" like printing to the console or writing to disk.
Note that not everything is an expression in F#. Type definitions are not expressions. Patterns are not expressions. And so on...
I found it's very hard to search for the simple indentation guide in F#.
Basically, I am wondering what's the rule for multiple-line statement indentation.
In C#, there is no problem because whitespace doesn't count.
Although I can write F# code according to my intuition and it works, I really want to know what's the rule for breaking one statement into multiple lines.
I write as
printfn "%d"
1
It works as expected
And if I write them in the same column, something goes wrong.
>
printfn "%A%A"
1
[];;
> //nothing is returned... and no error in this case
I want to confirm the basic rule for doing this. It's a little annoying when you can't be sure what you are doing.
Thanks in advance
I just tried another case
List.iter
(printfn "%d")
[1..10];;
And it prints out 1 to 10.
Why it's not
List.iter
((printfn "%d")
[1..10]);;
As Yin points out, the rule is that arguments of a function should be indented further than the call to the function. To add more details, your first snippet is interpreted like this:
printfn "%A%A";
1;
[];
Each of these is a valid expression that returns something (function, number, empty list) and then ignores the result and continues. Because they are written in the top-level scope, F# Interactive doesn't emit a warning that you're ignoring some values. If they were in a do block or let declaration:
do
printfn "%A%A"
1
[]
The F# compiler would emit a warning when sequencing expressions (using ;) that do not return unit:
stdin(5,3): warning FS0193: This expression is a function value, i.e. is missing arguments. Its type is 'a -> 'b -> unit.
stdin(6,3): warning FS0020: This expression should have type 'unit', but has typ
e 'int'. Use 'ignore' to discard the result of the expression, or 'let' to bind
the result to a name.
stdin(5,3): warning FS0020: This expression should have type 'unit', but has typ
e ''a list'. Use 'ignore' to discard the result of the expression, or 'let' to b
ind the result to a name.
In your second example, you should indent:
>
printfn "%A%A"
1
[];;
Otherwise the three expressions are three sequential expressions, not a single expression.
You can refer F# Language Specification for firm rules, e.g. Chapter 15 in the specification.
I've just found something I'd call a quirk in F# and would like to know whether it's by design or by mistake and if it's by design, why is it so...
If you write any range expression where the first term is greater than the second term the returned sequence is empty. A look at reflector suggests this is by design, but I can't really find a reason why it would have to be so.
An example to reproduce it is:
[1..10] |> List.length
[10..1] |> List.length
The first will print out 10 while the second will print out 0.
Tests were made in F# CTP 1.9.6.2.
EDIT: thanks for suggesting expliciting the range, but there's still one case (which is what inspired me to ask this question) that won't be covered. What if A and B are variables and none is constantly greater than the other although they're always different?
Considering that the range expression does not seem to get optimized at compiled time anyway, is there any good reason for the code which determines the step (not explicitly specified) in case A and B are ints not to allow negative steps?
As suggested by other answers, you can do
[10 .. -1 .. 1] |> List.iter (printfn "%A")
e.g.
[start .. step .. stop]
Adam Wright - But you should be able
to change the binding for types you're
interested in to behave in any way you
like (including counting down if x >
y).
Taking Adam's suggestion into code:
let (..) a b =
if a < b then seq { a .. b }
else seq { a .. -1 .. b }
printfn "%A" (seq { 1 .. 10 })
printfn "%A" (seq { 10 .. 1 })
This works for int ranges. Have a look at the source code for (..): you may be able to use that to work over other types of ranges, but not sure how you would get the right value of -1 for your specific type.
What "should" happen is, of course, subjective. Normal range notation in my mind defines [x..y] as the set of all elements greater than or equal to x AND less than or equal to y; an empty set if y < x. In this case, we need to appeal to the F# spec.
Range expressions expr1 .. expr2 are evaluated as a call to the overloaded operator (..), whose default binding is defined in Microsoft.FSharp.Core.Operators. This generates an IEnumerable<_> for the range of values between the given start (expr1) and finish (expr2) values, using an increment of 1. The operator requires the existence of a static member (..) (long name GetRange) on the static type of expr1 with an appropriate signature.
Range expressions expr1 .. expr2 .. expr3 are evaluated as a call to the overloaded operator (.. ..), whose default binding is defined in Microsoft.FSharp.Core.Operators. This generates an IEnumerable<_> for the range of values between the given start (expr1) and finish (expr3) values, using an increment of expr2. The operator requires the existence of a static member (..) (long name GetRange) on the static type of expr1 with an appropriate signature.
The standard doesn't seem to define the .. operator (a least, that I can find). But you should be able to change the binding for types you're interested in to behave in any way you like (including counting down if x > y).
In haskell, you can write [10, 9 .. 1]. Perhaps it works the same in F# (I haven't tried it)?
edit:
It seems that the F# syntax is different, maybe something like [10..-1..1]
Ranges are generally expressed (in the languages and frameworks that support them) like this:
low_value <to> high_value
Can you give a good argument why a range ought to be able to be expressed differently? Since you were requesting a range from a higher number to a lower number does it not stand to reason that the resulting range would have no members?