let (->>>) lst exp = for i in lst do exp
let result = [0..5] ->>> ([1..5] ->>> printf "hi")
let result2 = for i in [0..5] do for x in [1..5] do printf "hi"
I was expecting result and result2 to do the same thing. But result prints hi just once. result2, however, prints it 30 times. What is missing here?
In your operator definition, exp is not necessarily a function. The way you defined the operator, exp is any value whatsoever. For example, you could do this:
[0..1] ->>> 42
This would compile and even run, because there is nothing in the definition of your operator that requires the exp parameter to be a function. Or anything else for that matter.
If you want to make exp evaluate over and over within the body of the loop, you need to make it a function and make sure that the body of the loop calls it. Since you don't have any requirements for it, the simplest one would be unit -> 'a. You don't really have to declare its type though, you can have the compiler infer it from its use:
let (->>>) lst exp = for i in lst do exp()
Notice the parentheses after exp - they signify a value of type unit, thus making expression exp() a function application (i.e. function exp applied to value ()).
Of course, with this definition, you won't be able to write [1..5] ->>> printf "hi", because printf "hi" is not a function. So instead you'll have to write:
[1..5] ->>> (fun() -> printf "hi")
Now you'll have five "hi"s printed out.
This is because the for .. do syntax is really a special part of the language, not "just another function" defined in the standard library. You can't really create same effect with clever function or operator definitions.
If you want a way to construct more complicated syntax than functions and operators, check out computation expressions.
Related
The example code below appears to work nicely:
open FParsec
let capitalized : Parser<unit,unit> =(asciiUpper >>. many asciiLower >>. eof)
let inverted : Parser<unit,unit> =(asciiLower >>. many asciiUpper >>. eof)
let capsOrInvert =choice [capitalized;inverted]
You can then do:
run capsOrInvert "Dog";;
run capsOrInvert "dOG";;
and get a success or:
run capsOrInvert "dog";;
and get a failure.
Now that I have a ParserResult, how do I do things with it? For example, print the string backwards?
There are several notable issues with your code.
First off, as noticed in #scrwtp's answer, your parser returns unit. Here's why: operator (>>.) returns only the result returned by the right inner parser. On the other hand, (.>>) would return the result of a left parser, while (.>>.) would return a tuple of both left and right ones.
So, parser1 >>. parser2 >>. eof is essentially (parser1 >>. parser2) >>. eof.
The code in parens completely ignores the result of parser1, and the second (>>.) then ignores the entire result of the parser in parens. Finally, eof returns unit, and this value is being returned.
You may need some meaningful data returned instead, e.g. the parsed string. The easiest way is:
let capitalized = (asciiUpper .>>. many asciiLower .>> eof)
Mind the operators.
The code for inverted can be done in a similar manner.
This parser would be of type Parser<(char * char list), unit>, a tuple of first character and all the remaining ones, so you may need to merge them back. There are several ways to do that, here's one:
let mymerge (c1: char, cs: char list) = c1 :: cs // a simple cons
let pCapitalized = capitalized >>= mymerge
The beauty of this code is that your mymerge is a normal function, working with normal char's, it knows nothing about parsers or so. It just works with the data, and (>>=) operator does the rest.
Note, pCapitalized is also a parser, but it returns a single char list.
Nothing stops you from applying further transitions. As you mentioned printing the string backwards:
let pCapitalizedAndReversed =
capitalized
>>= mymerge
>>= List.rev
I have written the code in this way for purpose. In different lines you see a gradual transition of your domain data, still within the paradigm of Parser. This is an important consideration, because any subsequent transition may "decide" that the data is bad for some reason and raise a parsing exception, for example. Or, alternatively, it may be merged with other parser.
As soon as your domain data (a parsed-out word) is complete, you extract the result as mentioned in another answer.
A minor note. choice is superfluous for only two parsers. Use (<|>) instead. From experience, careful choosing parser combinators is important because a wrong choice deep inside your core parser logic can easily make your parsers dramatically slow.
See FParsec Primitives for further details.
ParserResult is a discriminated union. You simply match the Success and Failure cases.
let r = run capsOrInvert "Dog"
match r with
| Success(result, _, _) -> printfn "Success: %A" result
| Failure(errorMsg, _, _) -> printfn "Failure: %s" errorMsg
But this is probably not what you find tricky about your situation.
The thing about your Parser<unit, unit> type is that the parsed value is of type unit (the first type argument to Parser). What this means is that this parser doesn't really produce any sensible output for you to use - it can only tell you whether it can parse a string (in which case you get back a Success ((), _, _) - carrying the single value of type unit) or not.
What do you expect to get out of this parser?
Edit: This sounds close to what you want, or at least you should be able to pick up some pointers from it. capitalized accepts capitalized strings, inverted accepts capitalized strings that have been reversed and reverses them as part of the parser logic.
let reverse (s: string) =
System.String(Array.rev (Array.ofSeq s))
let capitalized : Parser<string,unit> =
(asciiUpper .>>. manyChars asciiLower)
|>> fun (upper, lower) -> string upper + lower
let inverted : Parser<string,unit> =
(manyChars asciiLower .>>. asciiUpper)
|>> fun (lower, upper) -> reverse (lower + string upper)
let capsOrInvert = choice [capitalized;inverted]
run capsOrInvert "Dog"
run capsOrInvert "doG"
run capsOrInvert "dog"
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.
How can I write a no-op statement in F#?
Specifically, how can I improve the second clause of the following match statement:
match list with
| [] -> printfn "Empty!"
| _ -> ignore 0
Use unit for empty side effect:
match list with
| [] -> printfn "Empty!"
| _ -> ()
The answer from Stringer is, of course, correct. I thought it may be useful to clarify how this works, because "()" insn't really an empty statement or empty side effect...
In F#, every valid piece of code is an expression. Constructs like let and match consist of some keywords, patterns and several sub-expressions. The F# grammar for let and match looks like this:
<expr> ::= let <pattern> = <expr>
<expr>
::= match <expr> with
| <pat> -> <expr>
This means that the body of let or the body of clause of match must be some expression. It can be some function call such as ignore 0 or it can be some value - in your case it must be some expression of type unit, because printfn ".." is also of type unit.
The unit type is a type that has only one value, which is written as () (and it also means empty tuple with no elements). This is, indeed, somewhat similar to void in C# with the exception that void doesn't have any values.
BTW: The following code may look like a sequence of statements, but it is also an expression:
printf "Hello "
printf "world"
The F# compiler implicitly adds ; between the two lines and ; is a sequencing operator, which has the following structure: <expr>; <expr>. It requires that the first expression returns unit and returns the result of the second expression.
This is a bit surprising when you're coming from C# background, but it makes the langauge surprisingly elegant and consise. It doesn't limit you in any way - you can for example write:
if (a < 10 && (printfn "demo"; true)) then // ...
(This example isn't really useful - just a demonstration of the flexibility)
I'm still working on a tiny parser for a tiny language defined in a task at school. The parser that generates an AST(Abstract syntax tree) is working. What I want is to check the defined variables, they must be bounded by the let expression. First the method that is defined in the task(suggestion, not needed):
checkVars :: Expr -> Char
data Expr = Var Char | Tall Int | Sum Expr Expr | Mult Expr Expr | Neg Expr | Let Expr Expr Expr
deriving(Eq, Show)
A valid sentence would be "let X be 5 in *(2,X)". X would normally be a Var and 5 is normally an int. And the last can be any part of the dataExpr type. Main point: X is used somewhere in the last expression. The datatype for let is:
Let Expr Expr Expr
Link to the other questions I've asked about this task here just FYI;
First question
Second question
As you see the datatype to the checkVars is Expr, so here is an example of what I would feed to that function:
parseProg "let X be 4 in let Y be *(2 , X) in let Z be +(Y , X) in
+(+(X , Y) , Z)"
Let (Var 'X') (Tall 4) (Let (Var 'Y') (Mult (Tall 2) (Var 'X')) (Let
(Var 'Z') (Sum (Var 'Y') (Var 'X')) (Sum (Sum (Var 'X') (Var 'Y')) (Var
'Z'))))
Just 24
This is an all-inclusive example, the top part is the string/program being parsed. The second part, starting at line 3 (Let) is the AST, input for the checkVars function. And the bottom part "Just 24" is the evaluation. Which I will be back here for more help for.
Note: The point is to spit out the first unbound variable found as an error, and ' ' if everything is fine. Obviously if you want to do this another way you can.
Here's something to think about:
The first field of your Let constructor is an Expr. But can it actually hold anything else than Vars? If not, you should reflect this by making that field's type, say, String and adapting the parser correspondingly. This will make your task a lot easier.
The standard trick to evaluating an expression with let-bindings (which you are doing) is to write a function
type Env = [(String, Int)]
eval :: Expr -> Env -> Int
Note the extra argument for the environment. The environment keeps track of what variables are bound at any given moment to what values. Its position in the type means that you get to decide its value every time you call eval on child expressions. This is crucial! It also means you can have locally declared variables: binding a variable has no effect on its context, only on subexpressions.
Here are the special cases:
In a Var, you want to lookup the variable name in the environment and return the value that is bound to it. (Use the standard Prelude function lookup.)
In a Let, you want to add an extra (varname, value) to the front of the environment list before passing it on to the child expression.
I've left out some details, but this should be enough to get you going a long way. If you get stuck, ask another question. :-)
Oh, and I see you want to return a Maybe value to indicate failure. I suggest you first try without and use error to indicate unbound variables. When you have that version of eval working, adapt it to return Maybe values. The reason for this is that working with Maybe values makes the evaluation quite a bit more complicated.
I would actually try to evaluate the AST. Start by processing (and thus removing) all the Lets. Now, try to evaluate the resulting AST. If you run across a Var then there is an unbound variable.