While studying for a Functional Programming exam, I came across the following question from a previous test:
t1 = (reverse . take 2 . words . \ _ -> name)"!"
The task is to write the output of the statement. The variable name refers to the student's name, written in the form "Smith, John". If I enter the statement into WinHugs, I get the following output:
["John","Smith,"]
I understand what the functions reverse, take and words are doing and I understand how the . operator connects them. What I don't understand is what is happening here:
\ _ -> name
What are the slash, underscore and "arrow" for? Also, what does the exclamation point in quotation marks do? (nothing?)
It's a lambda function that discards its (only) argument (i.e. "!") and yields name.
As another lambda example, the following would be a lambda function that squares its argument:
\x -> x * x
The \ is the notation used to introduce a lambda function.
The _ means "variable about whose name we do not care".
The -> separates the lambda function's arguments from the expression used to specify its result.
What you are seeing there is an anonymous function, or lambda function (that name comes from lambda calculus). The backslash tells you that you are starting the function. The underscore says that the function takes one argument and ignores it. The arrow points from the argument list to the result - in this case, it ends up ignoring its argument and returning the name. Essentially, \_ -> name is the same as const name.
A constant anonymous function: which ever the argument, return name.
Haskell's lambda expressions (i.e. anonymous functions) come in this form:
\x -> f x
where x is an argument, and f x an expression using this argument. The special variable _ matches anything and treats it as unimportant.
The "slash" is part of a lambda function, the underscore is a "wildcard" used in patterns (it is discarded). The arrow is another part of the lambda function. The function \ _ -> name returns the name, regardless of input, so the "!" does nothing but provide (unused) input to the function.
Related
I'm going through the book "Haskell programming from first principles" and although the authors try to explain in detail how to reduce expressions, I still have trouble understanding how to read some expressions in the examples provided.
Let's assume the below expression:
(λx.x) Then if I understand things correctly this can be read as: "A lambda function, that accepts an argument x and returns (basically translating . as "returns") the same value x.
Now supposing that that we have the below expression:
(λxy.xy), how do we read that expression in plain English? To make it even more complex the authors then expand (λxy.xy) to (λxy.xy)(λz.a) 1. How are these z and a appearing to the above expression? I was hoping that if I can properly transform the expression to plain English I'd be able to understand where z and a come from.
In lambda calculus, values that are next to each other imply that the first value is being called as a function with the second value as an argument. (λxy.xy) is shorthand for (λx.λy.xy), which means a function that takes an argument x then an argument y, and calls x with argument y. (if you're wondering why there's a function returning a function, it's due to currying). To figure out what (λxy.xy)(λz.a) 1 means, we have to take it step by step:
Undo the shorthand: (λx. λy. xy) (λz. a) 1
Apply first function (λx): (λy. (λz. a) y) 1
Apply second function (λy): (λz. a) 1
Apply third function (λz): a
So, we end up with whatever a is (probably an externally defined constant).
With GHC version 8.0.2 the following program:
import Debug.Trace
f=trace("f was called")$(+1)
main = do
print $ f 1
print $ f 2
outputs:
f was called
2
3
Is it the expected behaviour? If yes, why? I expected the string f was called to be printed twice, one before 2 and one before 3.
Same result on TIO: Try it online!
EDIT
But this program:
import Debug.Trace
f n=trace("f was called:"++show n)$n+1
main = do
print $ f 1
print $ f 2
outputs:
f was called:1
2
f was called:2
3
Try it online!
I suspect those behaviours have something to do with laziness, but my questions remain: is this the expected behaviour and, if yes, why?
Hackage asserts this:
The trace function outputs the trace message given as its first
argument, before returning the second argument as its result.
I don't see it in the first example.
EDIT 2 Third example based on #amalloy comments:
import Debug.Trace
f n=trace "f was called"$n+1
main = do
print $ f 1
print $ f 2
outputs:
f was called
2
f was called
3
Your trace prints when defining f, not when calling it. If you want the trace to happen as part of the call, you should make sure it is not evaluated until a parameter is received:
f x = trace "f was called" $ x + 1
Also, when I run your TIO I don't see the trace appearing at all. trace is not really a reliable way to print things, because it cheats the IO model that the language is built on. The most subtle changes in evaluation order can disturb it. Of course for debugging you can use it, but as even this simple example demonstrates it is not guaranteed to help much.
In your edit, you quote the documentation of trace:
The trace function outputs the trace message given as its first
argument, before returning the second argument as its result.
And indeed this is exactly what happens in your program! When defining f,
trace "f was called" $ (+ 1)
needs to be evaluated. First, "f was called" is printed. Then, trace evaluates to, and returns, (+ 1). This is the final value of the trace expression, and therefore (+ 1) is what f is defined as. The trace has vanished, see?
It is indeed a result of laziness.
Laziness means that merely defining a value doesn't mean it will be evaluated; that will only happen if it's needed for something. If it's not needed, the code that would actually produce it doesn't "do anything". If a particular value is needed the code is run, but only the first time it would be needed; if there are other references to the same value and it is used again, those uses will just directly use the value that was produced the first time.
You have to remember that functions are values in every sense of the term; everything that applies to ordinary values also applies to functions. So your definition of f is simply writing an expression for a value, the expression's evaluation will be deferred until the value of f is actually needed, and as it's needed twice the value (function) the expression computes will be saved and reused the second time.
Lets look at it in more detail:
f=trace("f was called")$(+1)
You're defining a value f with a simple equation (not using any syntactic sugar for writing arguments on the left hand side of the equation, or providing cases via multiple equations). So we can simply take the right hand side as a single expression that defines the value f. Just defining it does nothing, it sits there until you call:
print $ f 1
Now print needs its argument evaluated, so this is forcing the expression f 1. But we can't apply f to 1 without first forcing f. So we need to figure out what function the expression trace "f was called" $ (+1) evaluates to. So trace is actually called, does its unsafe IO printing and f was called appears at the terminal, and then trace returns its second argument: (+1).
So now we know what function f is: (+1). f will now be a direct reference to that function, with no need to evaluate the original code trace("f was called")$(+1) if f is called again. Which is why the second print does nothing.
This case is quite different, even though it might look similar:
f n=trace("f was called:"++show n)$n+1
Here we are using the syntactic sugar for defining functions by writing arguments on the left hand side. Let's desugar that to lambda notation to see more clearly what the actual value being bound to f is:
f = \n -> trace ("f was called:" ++ show n) $ n + 1
Here we've written a function value directly, rather than an expression that can be evaluated to result in a function. So when f needs to be evaluated before it can be called on 1, the value of f is that whole function; the trace call is inside the function instead of being the thing that is called to result in a function. So trace isn't called as part of evaluating f, it's called as part of evaluating the application f 1. If you saved the result of that (say by doing let x = f 1) and then printed it multiple times, you'd only see the one trace. But the when we come to evaluate f 2, the trace call is still there inside the function that is the value of f, so when f is called again so is trace.
I'm learning basic Haskell so I can configure Xmonad, and I ran into this code snippet:
newKeys x = myKeys x `M.union` keys def x
Now I understand what the M.union in backticks is and means. Here's how I'm interpreting it:
newKeys(x) = M.union(myKeys(x),???)
I don't know what to make of the keys def x. Is it like keys(def(x))? Or keys(def,x)? Or is def some sort of other keyword?
It's keys(def,x).
This is basic Haskell syntax for function application: first the function itself, then its arguments separated by spaces. For example:
f x y = x + y
z = f 5 6
-- z = 11
However, it is not clear what def is without larger context.
In response to your comment: no, def couldn't be a function that takes x as argument, and then the result of that is passed to keys. This is because function application is left-associative, which basically means that in any bunch of things separated by spaces, only the first one is the function being applied, and the rest are its arguments. In order to express keys(def(x)), one would have to write keys (def x).
If you want to be super technical, then the right way to think about it is that all functions have exactly one parameter. When we declare a function of two parameters, e.g. f x y = x + y, what we really mean is that it's a function of one parameter, which returns another function, to which we can then pass the remaining parameter. In other words, f 5 6 means (f 5) 6.
This idea is kind of one of the core things in Haskell (and any ML offshoot) syntax. It's so important that it has its own name - "currying" (after Haskell Curry, the mathematician).
For reference, here is my code: http://hpaste.org/86949
I am trying to parse the following expression: if (a[1].b[2].c.d[999].e[1+1].f > 3) { }. The method playing up is varExpr, which parses the variable member chains.
Context
In the language I am parsing, a dot can specify accessing a member variable. Since a member variable can be another object, chains can be produced ie: a.b.c, or essentially (a.b).c. Do not assume the dots are function composition.
Implementation
The logic is like this:
First, before <- many vocc collects all the instances of varname . and their optional array expression, leaving only a single identifier left
this <- vtrm collects the remaining identifier plus array expression -- the only one not proceeded by a dot
Issues
I am having two issues:
Firstly, the first term [for a reason that I cannot determine] seems to always require that it be wrapped in brackets for the parser to accept it ie: (a[1]).b[2].c... -- subsequent terms do not require this.
Secondly, the many vocc won't stop parsing. It always expects another identifier and another dot and I am unable to terminate the expression to catch the last vtrm.
I am looking for hints or solutions that will help me solve my problem(s)/headaches. Thanks.
When varExpr runs, it checks whether the next bit of input is matched by vocc or vtrm.
varExpr = do before <- many vocc -- Zero or more occurrences
this <- vtrm
return undefined
The problem is that any input matched by vtrm is also matched by the first step of vocc. When varExpr runs, it runs vocc, which runs vobj, which runs vtrm.
vocc = vobj <* symbol "."
vobj = choice [try vtrm, try $ parens vtrm]
Parsing of many vocc ends when vocc fails without consuming input. This happens when both vtrm and parens vtrm fail. However, after many vocc ends, the next parser to run is vtrm—and this parser is sure to fail!
You want vocc to fail without consuming input if it doesn't find a "." in the input. For that, you need to use try.
vocc = try $ vobj <* symbol "."
Alternatively, if vobj and vtrm really should be the same syntax, you can define varExpr as vobj `sepBy1` symbol ".".
I'm looking to call functions dynamically based on the contents found in an association list.
Here is an example in semi-pseudo-code. listOfFunctions would be passed to callFunctions.
listOfFunctions = [('function one', 'value one')
, ('function two', 'value two')
, ('function three', 'value three')]
callFunctions x = loop through functions
if entry found
then call function with value
else do nothing
The crux of the question is not looping through the list, rather, it's how to call a function once I have it's name?
Consider this use case for further clarification. You open the command prompt and are presented with the following menu.
1: Write new vHost file
2: Exit
You write the new vHost file and are not presented with a new menu
1: Enter new directive
2: Write file
3: Exit
You enter some new directives for the vHost and are now ready to write the file.
The program isn't going to blindly write each and every directive it can, rather, it will only write the ones that you supplied. This is where the association list comes in. Writing a giant if/then/else or case statement is madness. It would be much more elegant to loop through the list, look for which directives were added and call the functions to write them accordingly.
Hence, loop, find a function name, call said function with supplied value.
Thanks to anyone who can help out with this.
Edit:
Here is the solution that I've come up with (constructive critiques are always welcome).
I exported the functions which write the directives in an association list as every answer provided said that just including the function is the way to go.
funcMap = [("writeServerName", writeServerName)
,("writeServeralias", writeServerAlias)
,("writeDocRoot", writeDocRoot)
,("writeLogLevel", writeErrorLog)
,("writeErrorPipe", writeErrorPipe)
,("writeVhostOpen", writeVhostOpen)]
In the file which actually writes the hosts, that file is imported.
I have an association list called hostInfo to simulate some dummy value that would be gathered from an end-user and a function called runFunction which uses the technique supplied by edalorzo to filter through both the lists. By matching on the keys of both lists I ensure that the right function is called with the right value.
import Vhost.Directive
hostInfo = [("writeVhostOpen", "localhost:80")
,("writeServerName", "norics.com")]
runFunctions = [f val | (mapKey, f) <- funcMap, (key, val) <- hostInfo, mapKey == key]
You can simply include the function in the list directly; functions are values, so you can reference them by name in a list. Once you've got them out of the list, applying them is just as simple as func value. There's no need to involve their names at all.
Since I am farily new to Haskell I will risk that you consider my suggestion very naive, but anyways here it goes:
let funcs = [("sum", (+3),1),("product", (*3),2),("square", (^2),4)]
[f x | (name, f, x) <- funcs, name == "sum"]
I think it satisfies the requirements of the question, but perhaps what you intend is more sofisticated than what I can see with my yet limitted knowledge of Haskell.
It might be a bit of an overkill (I agree with ehird's reasoning) but you can evaluate a string with Haskell code by using the eval function in System.Eval.Haskell.
EDIT
As pointed out in the comments, hint is a better option for evaluating strings with Haskell expressions. Quoting the page:
This library defines an Interpreter monad. It allows to load Haskell modules, browse them, type-check and evaluate strings with Haskell expressions and even coerce them into values. The library is thread-safe and type-safe (even the coercion of expressions to values). It is, esentially, a huge subset of the GHC API wrapped in a simpler API. Works with GHC 6.10.x and 6.8.x
First we define our list of functions. This could be built using more machinery, but for the sake of example I just make one explicit list:
listOfFunctions :: [(Int, IO ())]
listOfFunctions = [(0, print "HI") -- notice the anonymous function
,(1, someNamedFunction) -- and something more traditional here
]
someNamedFunction = getChar >>= \x -> print x >> print x
Then we can select from this list however we want and execute the function:
executeFunctionWithVal :: Int -> IO ()
executeFunctionWithVal v = fromMaybe (return ()) (lookup v listOfFunctions)
and it works (if you import Data.Maybe):
Ok, modules loaded: Main.
> executeFunctionWithVal 0
"HI"
> executeFunctionWithVal 01
a'a'
'a'
Don't store the functions as strings, or rather, try storing the actual functions and then tagging them with a string. That way you can just call the function directly. Functions are first class values, so you can call the function using whatever name you assign it to.