I created a function which gets two string parameters. The function simply adds each string length. below is a code.
(defun add_twostring_length (mystr1 mystr2)
(+ (length mystr1) (length mystr2))
)
When I call add_twostring_length function like this,
(add_twostring_length "cpp" "lisp")
output is correct. 7
But, when I call the same function in the manner of using comma,
(add_twostring_length "cpp", "lisp")
I got an error message.
Error: Comma not inside a backquote.
[condition type: READER-ERROR]
I want to call function in the manner of (add_twostring_length "cpp", "lisp").
What is the wrong with the code?
picture showing error message
You might as well ask "why can't I call the function without parentheses?" In lisp, you call functions as an sexpr with the function in the car and the arguments in the cdr. There are no commas involved -- that's the syntax of lisp.
What you want is possible, but I will strongly advice against using it:
(set-macro-character #\,
#'(lambda (stream char)
(read stream t nil t)))
The above code creates the so called "read macro". At read-time common lisp will find all occurrences of , and ignore them. This makes possible calling functions like so:
(+ 1, 2, 3) ; => 6
However this will break escaping in templates:
`(1 2 ,(+ 3 4)) ; => (1 2 (+ 3 4))
Perhaps it is possible to make the read macro more intelligent, but I don't want to delve deeper into this, because I don't like the idea. Sorry.
I have a set of lambda expressions which I'm passing to other lambdas. All lambdas rely only on their arguments, they don't call any outside functions. Of course, sometimes it gets quite confusing and I'll pass an function with the incorrect number of arguments to another, creating a GHCi exception.
I want to make a debug function which will take an arbitrary lambda expression (with an unknown number of arguments) and return a string based on the structure and function of the lambda.
For example, say I have the following lambda expressions:
i = \x -> x
k = \x y -> x
s = \x y z -> x z (y z)
debug (s k) should return "\a b -> b"
debug (s s k) should return "\a b -> a b a" (if I simplified that correctly)
debug s should return "\a b c -> a c (b c)"
What would be a good way of doing this?
I think the way to do this would be to define a small lambda calculus DSL in Haskell (or use an existing implementation). This way, instead of using the native Haskell formulation, you would write something like
k = Lam "x" (Lam "y" (App (Var "x") (Var "y")))
s = Lam "x" (Lam "y" (Lam "z" (App (App (Var "x") (Var "z")
(App (Var "y") (Var "z"))))
and similarly for s and i. You would then write/use an evaluation function so that you could write
debug e = eval e
debug (App s k)
which would give you the final form in your own syntax. Additionally you would need a sort of interpreter to convert your DSL syntax to Haskell, so that you can actually use the functions in your code.
Implementing this does seem like quite a lot of (tricky) work, and it's probably not exactly what you had in mind (especially if you need the evaluation for typed syntax), but I'm sure it would be a great learning experience. A good reference would be chapter 6 of "Write you a Haskell". Using an existing implementation would be a lot easier (but less fun :)).
If this is merely for debugging purposes you might benefit from looking at the core syntax ghc compiles to. See chapter 25 of Real world Haskell, the ghc flag to use is -ddump-simpl. But this would mean looking at generated code rather than generating a representation inside your program. I'm also not sure to what extent you would be able to identify specific functions in the Core code easily (I have no experience with this so YMMV).
It would of course be pretty cool if using show on functions would give the kind of output you describe but there are probably very good reasons functions are not an instance of Show (I wouldn't be able to tell you).
You can actually achieve that by utilising pretty-printing from Template Haskell, which comes with GHC out of the box.
First, the formatting function should be defined in separate module (that's a TH restriction):
module LambdaPrint where
import Control.Monad
import Language.Haskell.TH.Ppr
import Language.Haskell.TH.Syntax
showDef :: Name -> Q Exp
showDef = liftM (LitE . StringL . pprint) . reify
Then use it:
{-# LANGUAGE TemplateHaskell #-}
import LambdaPrint
y :: a -> a
y = \a -> a
$(return []) --workaround for GHC 7.8+
test = $(showDef 'y)
The result is more or less readable, not counting fully qualified names:
*Main> test
"Main.y :: forall a_0 . a_0 -> a_0"
Few words about what's going on. showDef is a macro function which reifies the definition of some name from the environment and pretty-prints it in a string literal expression. To use it, you need to quote the name of the lambda (using ') and splice the result (which is a quoted string expression) into some expression (using $(...)).
I'm writing a compiler using uu-parsinglib and I saw a very strange thing. I defined a pChoice combinator like:
pChoice = foldr (<<|>) pFail
(notice, I'm using greedy <<|>).
Lets consider following code:
pFactor i = pChoice [ Expr.Var <$> pVar
, Expr.Lit <$> pLit True
, L.pParensed (pExpr i)
-- , Expr.Tuple <$> pTuple (pOpE i)
-- , Expr.List <$> pLst (pListE i)
]
Each element starts with different character - Expr.Var starts with a letter, Expr.Lit with a number, L.pParensed with parenthesis (, Expr.Tuple with brace { and Expr.List with bracket [.
I've got a big test code in which there are no tuples and no lists. The code parses in 0.15s. When I uncomment the above lines, the time increases to 0.65s. This is over 400% slowdown... How is it possible? I'm using only greedy operators and I'm sure parser is not haning in Tuple nor List section, because in the whole code there is no tuple nor list.
If you would need more code or definitions, I'll of course poste it.
I think the cause of the matter may lie in the fact that you have parameterised pFactor. This will cause each call to such a parser to build a new parser, which will take time. It is much better to create such parsers once and for all and share them in the actual parsing process. I cannot see how you are using this parser I cannot answer your questions any further.
ok, i have this problem here. i was asked to write a function in Scheme, that takes an "environment", and an expression and it returns the value of this expression, for the variable bindings found in the enviroment.
and a definition of a boolean expression is this according to the question below.
edit sorry my question is what does it mean by "it takes an environment" as an argument and what exactly does the function need to do?
evaluate for example "T OR F" and return "F" ???
"<expr> ::= <boolean>
|<variable>
|(not <expr>)
|(or <expr> <expr>)
|(and <expr> <expr>"
An environment is basically a dictionary of variable names to values. So given the environment
var1 = #t
var2 = #f
var3 = #t
and an expression
(or var2 (and T (or var1 var3)))
You would need to substitute the given values of var1, var2, and var3 into the expression, and then evaluate the expression.
Your function will probably be given the environment as some sort of Lisp structure, probably an alist, as one of its parameters.
From what I can determine you have been asked to implement eval.
So you should for starters have:
(define (eval expr env)
... )
where expr can be any of the forms you mention and env would keep the symbols defined (possibly a association list).
The first 2 cases are relatively trivial, the 3rd one, application of the not procedure should also be easy, given not will be in the environment (eg (list (cons 'not not))).
But the harder part lies in the last 2. Both of those are macros, and will require some expansion. The standard definitions/expansions of those should have been given to you. Once expanded, you can simply call eval recursively to evaluate the expanded expression.
Good luck :)
Edit:
Both and and or expands to if, so you will need to implement that too.
By "environment" they probably mean the equivalent of "scope" in other languages. Consider the following C fragment:
if (7 < 100)
{
int j = 2;
if (j < 4)
{
int k = 7, j = 14;
printf("k = %d, j = %d\n", k, j);
}
}
Note that in the outer scope (marked out by the outer set of braces) the only variable is j. In the inner scope there is a new j and a k. So there are three variables here, the outer j, and the inner j and k.
One way of implementing this is to define a scope to be a list of "environments". As you enter a new block, you put another "environment" in your list. When looking up variables by name, you look first in the most recently added "environment". If it isn't found there, you move along the list of environments to the next and look there, and so on.
An "environment" itself is often just a list of pairs, matching up names of variables with values. So it sounds like you are being asked to pass such a list to your function, each pair giving the symbol for a boolean variable and its value. Based on which variables are currently "in scope", you fetch their values out of the environment and use them in the expressions you are evaluating (according to that expression grammar you've been given).
In your case, it sounds like you aren't being asked to worry about which enviroments are in scope. You just have one environment, i.e. one list of pairs.
Sounds like a fair bit of work, good luck!
One reference that might help is:
http://michaux.ca/articles/scheme-from-scratch-bootstrap-v0_9-environments
What are the precise rules for when you can omit (omit) parentheses, dots, braces, = (functions), etc.?
For example,
(service.findAllPresentations.get.first.votes.size) must be equalTo(2).
service is my object
def findAllPresentations: Option[List[Presentation]]
votes returns List[Vote]
must and be are both functions of specs
Why can't I go:
(service findAllPresentations get first votes size) must be equalTo(2)
?
The compiler error is:
"RestServicesSpecTest.this.service.findAllPresentations
of type
Option[List[com.sharca.Presentation]]
does not take parameters"
Why does it think I'm trying to pass in a parameter? Why must I use dots for every method call?
Why must (service.findAllPresentations get first votes size) be equalTo(2) result in:
"not found: value first"
Yet, the "must be equalTo 2" of
(service.findAllPresentations.get.first.votes.size) must be equalTo 2, that is, method chaining works fine? - object chain chain chain param.
I've looked through the Scala book and website and can't really find a comprehensive explanation.
Is it in fact, as Rob H explains in Stack Overflow question Which characters can I omit in Scala?, that the only valid use-case for omitting the '.' is for "operand operator operand" style operations, and not for method chaining?
You seem to have stumbled upon the answer. Anyway, I'll try to make it clear.
You can omit dot when using the prefix, infix and postfix notations -- the so called operator notation. While using the operator notation, and only then, you can omit the parenthesis if there is less than two parameters passed to the method.
Now, the operator notation is a notation for method-call, which means it can't be used in the absence of the object which is being called.
I'll briefly detail the notations.
Prefix:
Only ~, !, + and - can be used in prefix notation. This is the notation you are using when you write !flag or val liability = -debt.
Infix:
That's the notation where the method appears between an object and it's parameters. The arithmetic operators all fit here.
Postfix (also suffix):
That notation is used when the method follows an object and receives no parameters. For example, you can write list tail, and that's postfix notation.
You can chain infix notation calls without problem, as long as no method is curried. For example, I like to use the following style:
(list
filter (...)
map (...)
mkString ", "
)
That's the same thing as:
list filter (...) map (...) mkString ", "
Now, why am I using parenthesis here, if filter and map take a single parameter? It's because I'm passing anonymous functions to them. I can't mix anonymous functions definitions with infix style because I need a boundary for the end of my anonymous function. Also, the parameter definition of the anonymous function might be interpreted as the last parameter to the infix method.
You can use infix with multiple parameters:
string substring (start, end) map (_ toInt) mkString ("<", ", ", ">")
Curried functions are hard to use with infix notation. The folding functions are a clear example of that:
(0 /: list) ((cnt, string) => cnt + string.size)
(list foldLeft 0) ((cnt, string) => cnt + string.size)
You need to use parenthesis outside the infix call. I'm not sure the exact rules at play here.
Now, let's talk about postfix. Postfix can be hard to use, because it can never be used anywhere except the end of an expression. For example, you can't do the following:
list tail map (...)
Because tail does not appear at the end of the expression. You can't do this either:
list tail length
You could use infix notation by using parenthesis to mark end of expressions:
(list tail) map (...)
(list tail) length
Note that postfix notation is discouraged because it may be unsafe.
I hope this has cleared all the doubts. If not, just drop a comment and I'll see what I can do to improve it.
Class definitions:
val or var can be omitted from class parameters which will make the parameter private.
Adding var or val will cause it to be public (that is, method accessors and mutators are generated).
{} can be omitted if the class has no body, that is,
class EmptyClass
Class instantiation:
Generic parameters can be omitted if they can be inferred by the compiler. However note, if your types don't match, then the type parameter is always infered so that it matches. So without specifying the type, you may not get what you expect - that is, given
class D[T](val x:T, val y:T);
This will give you a type error (Int found, expected String)
var zz = new D[String]("Hi1", 1) // type error
Whereas this works fine:
var z = new D("Hi1", 1)
== D{def x: Any; def y: Any}
Because the type parameter, T, is inferred as the least common supertype of the two - Any.
Function definitions:
= can be dropped if the function returns Unit (nothing).
{} for the function body can be dropped if the function is a single statement, but only if the statement returns a value (you need the = sign), that is,
def returnAString = "Hi!"
but this doesn't work:
def returnAString "Hi!" // Compile error - '=' expected but string literal found."
The return type of the function can be omitted if it can be inferred (a recursive method must have its return type specified).
() can be dropped if the function doesn't take any arguments, that is,
def endOfString {
return "myDog".substring(2,1)
}
which by convention is reserved for methods which have no side effects - more on that later.
() isn't actually dropped per se when defining a pass by name paramenter, but it is actually a quite semantically different notation, that is,
def myOp(passByNameString: => String)
Says myOp takes a pass-by-name parameter, which results in a String (that is, it can be a code block which returns a string) as opposed to function parameters,
def myOp(functionParam: () => String)
which says myOp takes a function which has zero parameters and returns a String.
(Mind you, pass-by-name parameters get compiled into functions; it just makes the syntax nicer.)
() can be dropped in the function parameter definition if the function only takes one argument, for example:
def myOp2(passByNameString:(Int) => String) { .. } // - You can drop the ()
def myOp2(passByNameString:Int => String) { .. }
But if it takes more than one argument, you must include the ():
def myOp2(passByNameString:(Int, String) => String) { .. }
Statements:
. can be dropped to use operator notation, which can only be used for infix operators (operators of methods that take arguments). See Daniel's answer for more information.
. can also be dropped for postfix functions
list tail
() can be dropped for postfix operators
list.tail
() cannot be used with methods defined as:
def aMethod = "hi!" // Missing () on method definition
aMethod // Works
aMethod() // Compile error when calling method
Because this notation is reserved by convention for methods that have no side effects, like List#tail (that is, the invocation of a function with no side effects means that the function has no observable effect, except for its return value).
() can be dropped for operator notation when passing in a single argument
() may be required to use postfix operators which aren't at the end of a statement
() may be required to designate nested statements, ends of anonymous functions or for operators which take more than one parameter
When calling a function which takes a function, you cannot omit the () from the inner function definition, for example:
def myOp3(paramFunc0:() => String) {
println(paramFunc0)
}
myOp3(() => "myop3") // Works
myOp3(=> "myop3") // Doesn't work
When calling a function that takes a by-name parameter, you cannot specify the argument as a parameter-less anonymous function. For example, given:
def myOp2(passByNameString:Int => String) {
println(passByNameString)
}
You must call it as:
myOp("myop3")
or
myOp({
val source = sourceProvider.source
val p = myObject.findNameFromSource(source)
p
})
but not:
myOp(() => "myop3") // Doesn't work
IMO, overuse of dropping return types can be harmful for code to be re-used. Just look at specification for a good example of reduced readability due to lack of explicit information in the code. The number of levels of indirection to actually figure out what the type of a variable is can be nuts. Hopefully better tools can avert this problem and keep our code concise.
(OK, in the quest to compile a more complete, concise answer (if I've missed anything, or gotten something wrong/inaccurate please comment), I have added to the beginning of the answer. Please note this isn't a language specification, so I'm not trying to make it exactly academically correct - just more like a reference card.)
A collection of quotes giving insight into the various conditions...
Personally, I thought there'd be more in the specification. I'm sure there must be, I'm just not searching for the right words...
There are a couple of sources however, and I've collected them together, but nothing really complete / comprehensive / understandable / that explains the above problems to me...:
"If a method body has more than one
expression, you must surround it with
curly braces {…}. You can omit the
braces if the method body has just one
expression."
From chapter 2, "Type Less, Do More", of Programming Scala:
"The body of the upper method comes
after the equals sign ‘=’. Why an
equals sign? Why not just curly braces
{…}, like in Java? Because semicolons,
function return types, method
arguments lists, and even the curly
braces are sometimes omitted, using an
equals sign prevents several possible
parsing ambiguities. Using an equals
sign also reminds us that even
functions are values in Scala, which
is consistent with Scala’s support of
functional programming, described in
more detail in Chapter 8, Functional
Programming in Scala."
From chapter 1, "Zero to Sixty: Introducing Scala", of Programming Scala:
"A function with no parameters can be
declared without parentheses, in which
case it must be called with no
parentheses. This provides support for
the Uniform Access Principle, such
that the caller does not know if the
symbol is a variable or a function
with no parameters.
The function body is preceded by "="
if it returns a value (i.e. the return
type is something other than Unit),
but the return type and the "=" can be
omitted when the type is Unit (i.e. it
looks like a procedure as opposed to a
function).
Braces around the body are not
required (if the body is a single
expression); more precisely, the body
of a function is just an expression,
and any expression with multiple parts
must be enclosed in braces (an
expression with one part may
optionally be enclosed in braces)."
"Functions with zero or one argument
can be called without the dot and
parentheses. But any expression can
have parentheses around it, so you can
omit the dot and still use
parentheses.
And since you can use braces anywhere
you can use parentheses, you can omit
the dot and put in braces, which can
contain multiple statements.
Functions with no arguments can be
called without the parentheses. For
example, the length() function on
String can be invoked as "abc".length
rather than "abc".length(). If the
function is a Scala function defined
without parentheses, then the function
must be called without parentheses.
By convention, functions with no
arguments that have side effects, such
as println, are called with
parentheses; those without side
effects are called without
parentheses."
From blog post Scala Syntax Primer:
"A procedure definition is a function
definition where the result type and
the equals sign are omitted; its
defining expression must be a block.
E.g., def f (ps) {stats} is
equivalent to def f (ps): Unit =
{stats}.
Example 4.6.3 Here is a declaration
and a de?nition of a procedure named
write:
trait Writer {
def write(str: String)
}
object Terminal extends Writer {
def write(str: String) { System.out.println(str) }
}
The code above is implicitly completed
to the following code:
trait Writer {
def write(str: String): Unit
}
object Terminal extends Writer {
def write(str: String): Unit = { System.out.println(str) }
}"
From the language specification:
"With methods which only take a single
parameter, Scala allows the developer
to replace the . with a space and omit
the parentheses, enabling the operator
syntax shown in our insertion operator
example. This syntax is used in other
places in the Scala API, such as
constructing Range instances:
val firstTen:Range = 0 to 9
Here again, to(Int) is a vanilla
method declared inside a class
(there’s actually some more implicit
type conversions here, but you get the
drift)."
From Scala for Java Refugees Part 6: Getting Over Java:
"Now, when you try "m 0", Scala
discards it being a unary operator, on
the grounds of not being a valid one
(~, !, - and +). It finds that "m" is
a valid object -- it is a function,
not a method, and all functions are
objects.
As "0" is not a valid Scala
identifier, it cannot be neither an
infix nor a postfix operator.
Therefore, Scala complains that it
expected ";" -- which would separate
two (almost) valid expressions: "m"
and "0". If you inserted it, then it
would complain that m requires either
an argument, or, failing that, a "_"
to turn it into a partially applied
function."
"I believe the operator syntax style
works only when you've got an explicit
object on the left-hand side. The
syntax is intended to let you express
"operand operator operand" style
operations in a natural way."
Which characters can I omit in Scala?
But what also confuses me is this quote:
"There needs to be an object to
receive a method call. For instance,
you cannot do “println “Hello World!”"
as the println needs an object
recipient. You can do “Console
println “Hello World!”" which
satisfies the need."
Because as far as I can see, there is an object to receive the call...
I find it easier to follow this rule of thumb: in expressions spaces alternate between methods and parameters. In your example, (service.findAllPresentations.get.first.votes.size) must be equalTo(2) parses as (service.findAllPresentations.get.first.votes.size).must(be)(equalTo(2)). Note that the parentheses around the 2 have a higher associativity than the spaces. Dots also have higher associativity, so (service.findAllPresentations.get.first.votes.size) must be.equalTo(2)would parse as (service.findAllPresentations.get.first.votes.size).must(be.equalTo(2)).
service findAllPresentations get first votes size must be equalTo 2 parses as service.findAllPresentations(get).first(votes).size(must).be(equalTo).2.
Actually, on second reading, maybe this is the key:
With methods which only take a single
parameter, Scala allows the developer
to replace the . with a space and omit
the parentheses
As mentioned on the blog post: http://www.codecommit.com/blog/scala/scala-for-java-refugees-part-6 .
So perhaps this is actually a very strict "syntax sugar" which only works where you are effectively calling a method, on an object, which takes one parameter. e.g.
1 + 2
1.+(2)
And nothing else.
This would explain my examples in the question.
But as I said, if someone could point out to be exactly where in the language spec this is specified, would be great appreciated.
Ok, some nice fellow (paulp_ from #scala) has pointed out where in the language spec this information is:
6.12.3:
Precedence and associativity of
operators determine the grouping of
parts of an expression as follows.
If there are several infix operations in an expression, then
operators with higher precedence bind
more closely than operators with lower
precedence.
If there are consecutive infix operations e0 op1 e1 op2 . . .opn en
with operators op1, . . . , opn of the
same precedence, then all these
operators must have the same
associativity. If all operators are
left-associative, the sequence is
interpreted as (. . . (e0 op1 e1) op2
. . .) opn en. Otherwise, if all
operators are rightassociative, the
sequence is interpreted as e0 op1 (e1
op2 (. . .opn en) . . .).
Postfix operators always have lower precedence than infix operators. E.g.
e1 op1 e2 op2 is always equivalent to
(e1 op1 e2) op2.
The right-hand operand of a
left-associative operator may consist
of several arguments enclosed in
parentheses, e.g. e op (e1, . . .
,en). This expression is then
interpreted as e.op(e1, . . . ,en).
A left-associative binary operation e1
op e2 is interpreted as e1.op(e2). If
op is rightassociative, the same
operation is interpreted as { val
x=e1; e2.op(x ) }, where x is a fresh
name.
Hmm - to me it doesn't mesh with what I'm seeing or I just don't understand it ;)
There aren't any. You will likely receive advice around whether or not the function has side-effects. This is bogus. The correction is to not use side-effects to the reasonable extent permitted by Scala. To the extent that it cannot, then all bets are off. All bets. Using parentheses is an element of the set "all" and is superfluous. It does not provide any value once all bets are off.
This advice is essentially an attempt at an effect system that fails (not to be confused with: is less useful than other effect systems).
Try not to side-effect. After that, accept that all bets are off. Hiding behind a de facto syntactic notation for an effect system can and does, only cause harm.