This is what Rich Hickey said in one of the blog posts but I don't understand the motivation in using apply. Please help.
A big difference between Clojure and CL is that Clojure is a Lisp-1, so funcall is not needed, and apply is only used to apply a function to a runtime-defined collection of arguments. So, (apply f [i]) can be written (f i).
Also, what does he mean by "Clojure is Lisp-1" and funcall is not needed? I have never programmed in CL.
Thanks
You would use apply, if the number of arguments to pass to the function is not known at compile-time (sorry, don't know Clojure syntax all that well, resorting to Scheme):
(define (call-other-1 func arg) (func arg))
(define (call-other-2 func arg1 arg2) (func arg1 arg2))
As long as the number of arguments is known at compile time, you can pass them directly as is done in the example above. But if the number of arguments is not known at compile-time, you cannot do this (well, you could try something like):
(define (call-other-n func . args)
(case (length args)
((0) (other))
((1) (other (car args)))
((2) (other (car args) (cadr args)))
...))
but that becomes a nightmare soon enough. That's where apply enters the picture:
(define (call-other-n func . args)
(apply other args))
It takes whatever number of arguments are contained in the list given as last argument to it, and calls the function passed as first argument to apply with those values.
The terms Lisp-1 and Lisp-2 refer to whether functions are in the same namespace as variables.
In a Lisp-2 (that is, 2 namespaces), the first item in a form will be evaluated as a function name — even if it's actually the name of a variable with a function value. So if you want to call a variable function, you have to pass the variable to another function.
In a Lisp-1, like Scheme and Clojure, variables that evaluate to functions can go in the initial position, so you don't need to use apply in order to evaluate it as a function.
apply basically unwraps a sequence and applies the function to them as individual arguments.
Here is an example:
(apply + [1 2 3 4 5])
That returns 15. It basically expands to (+ 1 2 3 4 5), instead of (+ [1 2 3 4 5]).
You use apply to convert a function that works on several arguments to one that works on a single sequence of arguments. You can also insert arguments before the sequence. For example, map can work on several sequences. This example (from ClojureDocs) uses map to transpose a matrix.
user=> (apply map vector [[:a :b] [:c :d]])
([:a :c] [:b :d])
The one inserted argument here is vector. So the apply expands to
user=> (map vector [:a :b] [:c :d])
Cute!
PS To return a vector of vectors instead of a sequence of vectors, wrap the whole thing in vec:
user=> (vec (apply map vector [[:a :b] [:c :d]]))
While we're here, vec could be defined as (partial apply vector), though it isn't.
Concerning Lisp-1 and Lisp-2: the 1 and 2 indicate the number of things a name can denote in a given context. In a Lisp-2, you can have two different things (a function and a variable) with the same name. So, wherever either might be valid, you need to decorate your program with something to indicate which you mean. Thankfully, Clojure (or Scheme ...) allows a name to denote just one thing, so no such decorations are necessary.
The usual pattern for apply type operations is to combine a function provided at runtime with a set of arguments, ditto.
I've not done enough with clojure to be able to be confident about the subtleties for that particular language to tell whether the use of apply in that case would be strictly necessary.
Apply is useful with protocols, especially in conjunction with threading macros. I just discovered this. Since you can't use the & macro to expand interface arguments at compile time, you can apply an unpredictably sized vector instead.
So I use this, for instance, as part of an interface between a record holding some metadata about a particular xml file and the file itself.
(query-tree [this forms]
(apply xml-> (text-id-to-tree this) forms)))
text-id-to-tree is another method of this particular record that parses a file into an xml zipper. In another file, I extend the protocol with a particular query that implements query-tree, specifying a chain of commands to be threaded through the xml-> macro:
(tags-with-attrs [this]
(query-tree this [zf/descendants zip/node (fn [node] [(map #(% node) [:tag :attrs])])])
(note: this query by itself will return a lot of "nil" results for tags that don't have
attributes. Filter and reduce for a clean list of unique values).
zf, by the way, refers to clojure.contrib.zip-filter, and zip to clojure.zip. The xml-> macro is from the clojure.contrib.zip-filter.xml library, which I :use
Related
I am trying to do Recursive method to find max value in list.
Can anyone explain where I made the mistake on this code and how to approach it next time.
(defun f3 (i)
(setq x (cond (> (car (I)) (cdr (car (I))))
(f3 (cdr (I)))))
)
(f3 '(33 11 44 2) )
also I tried this following method and didn't work:
(defun f3 (i)
(cond ((null I )nil )
(setq x (car (i))
(f3(cdr (i)))
(return-from max x)
)
Thanks a lot for any help. I am coming from java if that helps.
If you're working in Common Lisp, then you do this:
(defun max-item (list)
(loop for item in list
maximizing item))
That's it. The maximizing item clause of loop determines the highest item value seen, and implicitly establishes that as the result value of loop when it terminates.
Note that if list is empty, then this returns nil. If you want some other behavior, you have to work that in:
(if list
(loop for item in list
maximizing item))
(... handle empty here ...))
If the number of elements in the list is known to be small, below your Lisp implementation's limit on the number of arguments that can be passed to a function, you can simply apply the list to the max function:
(defun max-item (list)
(apply #'max list))
If list is empty, then max is misused: it requires one or more arguments. An error condition will likely be signaled. If that doesn't work in your situation, you need to add code to supply the desired behavior.
If the list is expected to be large, so that this approach is to be avoided, you can use reduce, treating max as a binary function:
(defun max-item (list)
(reduce #'max list))
Same remarks regarding empty list. These expressions are so small, many programmers will avoid writing a function and just use them directly.
Regarding recursion, you wouldn't use recursion to solve this problem in production code, only as a homework exercise for learning about recursion.
You are trying to compute the maximum value of a list, so please name your function maximum and your parameter list, not f3 or i. You can't name the function max without having to consider how to avoid shadowing the standard max function, so it is best for now to ignore package issues and use a different name.
There is a corner case to consider when the list is empty, as there is no meaningful value to return. You have to decide if you return nil or signal an error, for example.
The skeleton is thus:
(defun maximum (list)
(if (null list)
...
...))
Notice how closing parentheses are never preceded by spaces (or newlines), and opening parentheses are never followed by spaces (or newlines). Please note also that indentation increases with the current depth . This is the basic rules for Lisp formatting, please try following them for other developers.
(setq x <value>)
You are assigning an unknown place x, you should instead bind a fresh variable if you want to have a temporary variable, something like:
(let ((x <value>))
<body>)
With the above expression, x is bound to <value> inside <body> (one or more expressions), and only there.
(car (i))
Unlike in Java, parentheses are not used to group expressions for readability or to force some evaluation order, in Lisp they enclose compound forms. Here above, in a normal evaluation context (not a macro or binding), (i) means call function i, and this function is unrelated to your local variable i (just like in Java, where you can write int f = f(2) with f denoting both a variable and a method).
If you want to take the car of i, write (car i).
You seem to be using cond as some kind of if:
(cond (<test> <then>) <else>) ;; WRONG
You can have an if as follows:
(if <test> <then> <else>)
For example:
(if (> u v) u v) ;; evaluates to either `u` or `v`, whichever is greater
The cond syntax is a bit more complex but you don't need it yet.
You cannot return-from a block that was undeclared, you probably renamed the function to f3 without renaming that part, or copied that from somewhere else, but in any case return-from is only needed when you have a bigger function and probably a lot more side-effects. Here the computation can be written in a more functionnal way. There is an implicit return in Lisp-like languages, unlike Java, for example below the last (but also single) expression in add evaluates to the function's return value:
(defun add-3 (x)
(+ x 3))
Start with smaller examples and test often, fix any error the compiler or interpreter prints before trying to do more complex things. Also, have a look at the available online resources to learn more about the language: https://common-lisp.net/documentation
Although the other answers are right: you definitely need to learn more CL syntax and you probably would not solve this problem recursively in idiomatic CL (whatever 'idiomatic CL' is), here's how to actually do it, because thinking about how to solve these problems recursively is useful.
First of all let's write a function max/2 which returns the maximum of two numbers. This is pretty easy:
(defun max/2 (a b)
(if (> a b) a b))
Now the trick is this: assume you have some idea of what the maximum of a list of numbers is: call this guess m. Then:
if the list is empty, the maximum is m;
otherwise the list has a first element, so pick a new m which is the maximum of the first element of the list and the current m, and recurse on the rest of the list.
So, we can write this function, which I'll call max/carrying (because it 'carries' the m):
(defun max/carrying (m list)
(if (null list)
m
(max/carrying (max/2 (first list) m)
(rest list))))
And this is now almost all we need. The trick is then to write a little shim around max/carrying which bootstraps it:
to compute the maximum of a list:
if the list is empty it has no maximum, and this is an error;
otherwise the result is max/carrying of the first element of the list and the rest of the list.
I won't write that, but it's pretty easy (to signal an error, the function you want is error).
After making it through the major parts of an introductory Lisp book, I still couldn't understand what the special operator (quote) (or equivalent ') function does, yet this has been all over Lisp code that I've seen.
What does it do?
Short answer
Bypass the default evaluation rules and do not evaluate the expression (symbol or s-exp), passing it along to the function exactly as typed.
Long Answer: The Default Evaluation Rule
When a regular (I'll come to that later) function is invoked, all arguments passed to it are evaluated. This means you can write this:
(* (+ a 2)
3)
Which in turn evaluates (+ a 2), by evaluating a and 2. The value of the symbol a is looked up in the current variable binding set, and then replaced. Say a is currently bound to the value 3:
(let ((a 3))
(* (+ a 2)
3))
We'd get (+ 3 2), + is then invoked on 3 and 2 yielding 5. Our original form is now (* 5 3) yielding 15.
Explain quote Already!
Alright. As seen above, all arguments to a function are evaluated, so if you would like to pass the symbol a and not its value, you don't want to evaluate it. Lisp symbols can double both as their values, and markers where you in other languages would have used strings, such as keys to hash tables.
This is where quote comes in. Say you want to plot resource allocations from a Python application, but rather do the plotting in Lisp. Have your Python app do something like this:
print("'(")
while allocating:
if random.random() > 0.5:
print(f"(allocate {random.randint(0, 20)})")
else:
print(f"(free {random.randint(0, 20)})")
...
print(")")
Giving you output looking like this (slightly prettyfied):
'((allocate 3)
(allocate 7)
(free 14)
(allocate 19)
...)
Remember what I said about quote ("tick") causing the default rule not to apply? Good. What would otherwise happen is that the values of allocate and free are looked up, and we don't want that. In our Lisp, we wish to do:
(dolist (entry allocation-log)
(case (first entry)
(allocate (plot-allocation (second entry)))
(free (plot-free (second entry)))))
For the data given above, the following sequence of function calls would have been made:
(plot-allocation 3)
(plot-allocation 7)
(plot-free 14)
(plot-allocation 19)
But What About list?
Well, sometimes you do want to evaluate the arguments. Say you have a nifty function manipulating a number and a string and returning a list of the resulting ... things. Let's make a false start:
(defun mess-with (number string)
'(value-of-number (1+ number) something-with-string (length string)))
Lisp> (mess-with 20 "foo")
(VALUE-OF-NUMBER (1+ NUMBER) SOMETHING-WITH-STRING (LENGTH STRING))
Hey! That's not what we wanted. We want to selectively evaluate some arguments, and leave the others as symbols. Try #2!
(defun mess-with (number string)
(list 'value-of-number (1+ number) 'something-with-string (length string)))
Lisp> (mess-with 20 "foo")
(VALUE-OF-NUMBER 21 SOMETHING-WITH-STRING 3)
Not Just quote, But backquote
Much better! Incidently, this pattern is so common in (mostly) macros, that there is special syntax for doing just that. The backquote:
(defun mess-with (number string)
`(value-of-number ,(1+ number) something-with-string ,(length string)))
It's like using quote, but with the option to explicitly evaluate some arguments by prefixing them with comma. The result is equivalent to using list, but if you're generating code from a macro you often only want to evaluate small parts of the code returned, so the backquote is more suited. For shorter lists, list can be more readable.
Hey, You Forgot About quote!
So, where does this leave us? Oh right, what does quote actually do? It simply returns its argument(s) unevaluated! Remember what I said in the beginning about regular functions? Turns out that some operators/functions need to not evaluate their arguments. Such as IF -- you wouldn't want the else branch to be evaluated if it wasn't taken, right? So-called special operators, together with macros, work like that. Special operators are also the "axiom" of the language -- minimal set of rules -- upon which you can implement the rest of Lisp by combining them together in different ways.
Back to quote, though:
Lisp> (quote spiffy-symbol)
SPIFFY-SYMBOL
Lisp> 'spiffy-symbol ; ' is just a shorthand ("reader macro"), as shown above
SPIFFY-SYMBOL
Compare to (on Steel-Bank Common Lisp):
Lisp> spiffy-symbol
debugger invoked on a UNBOUND-VARIABLE in thread #<THREAD "initial thread" RUNNING {A69F6A9}>:
The variable SPIFFY-SYMBOL is unbound.
Type HELP for debugger help, or (SB-EXT:QUIT) to exit from SBCL.
restarts (invokable by number or by possibly-abbreviated name):
0: [ABORT] Exit debugger, returning to top level.
(SB-INT:SIMPLE-EVAL-IN-LEXENV SPIFFY-SYMBOL #<NULL-LEXENV>)
0]
Because there is no spiffy-symbol in the current scope!
Summing Up
quote, backquote (with comma), and list are some of the tools you use to create lists, that are not only lists of values, but as you seen can be used as lightweight (no need to define a struct) data structures!
If you wish to learn more, I recommend Peter Seibel's book Practical Common Lisp for a practical approach to learning Lisp, if you're already into programming at large. Eventually on your Lisp journey, you'll start using packages too. Ron Garret's The Idiot's Guide to Common Lisp Packages will give you good explanation of those.
Happy hacking!
It says "don't evaluate me". For example, if you wanted to use a list as data, and not as code, you'd put a quote in front of it. For example,
(print '(+ 3 4)) prints "(+ 3 4)", whereas
(print (+ 3 4)) prints "7"
Other people have answered this question admirably, and Matthias Benkard brings up an excellent warning.
DO NOT USE QUOTE TO CREATE LISTS THAT YOU WILL LATER MODIFY. The spec allows the compiler to treat quoted lists as constants. Often, a compiler will optimize constants by creating a single value for them in memory and then referencing that single value from all locations where the constant appears. In other words, it may treat the constant like an anonymous global variable.
This can cause obvious problems. If you modify a constant, it may very well modify other uses of the same constant in completely unrelated code. For example, you may compare some variable to '(1 1) in some function, and in a completely different function, start a list with '(1 1) and then add more stuff to it. Upon running these functions, you may find that the first function doesn't match things properly anymore, because it's now trying to compare the variable to '(1 1 2 3 5 8 13), which is what the second function returned. These two functions are completely unrelated, but they have an effect on each other because of the use of constants. Even crazier bad effects can happen, like a perfectly normal list iteration suddenly infinite looping.
Use quote when you need a constant list, such as for comparison. Use list when you will be modifying the result.
One answer to this question says that QUOTE “creates list data structures”. This isn't quite right. QUOTE is more fundamental than this. In fact, QUOTE is a trivial operator: Its purpose is to prevent anything from happening at all. In particular, it doesn't create anything.
What (QUOTE X) says is basically “don't do anything, just give me X.” X needn't be a list as in (QUOTE (A B C)) or a symbol as in (QUOTE FOO). It can be any object whatever. Indeed, the result of evaluating the list that is produced by (LIST 'QUOTE SOME-OBJECT) will always just return SOME-OBJECT, whatever it is.
Now, the reason that (QUOTE (A B C)) seems as if it created a list whose elements are A, B, and C is that such a list really is what it returns; but at the time the QUOTE form is evaluated, the list has generally already been in existence for a while (as a component of the QUOTE form!), created either by the loader or the reader prior to execution of the code.
One implication of this that tends to trip up newbies fairly often is that it's very unwise to modify a list returned by a QUOTE form. Data returned by QUOTE is, for all intents and purposes, to be considered as part of the code being executed and should therefore be treated as read-only!
The quote prevents execution or evaluation of a form, turning it instead into data. In general you can execute the data by then eval'ing it.
quote creates list data structures, for example, the following are equivalent:
(quote a)
'a
It can also be used to create lists (or trees):
(quote (1 2 3))
'(1 2 3)
You're probably best off getting an introductary book on lisp, such as Practical Common Lisp (which is available to read on-line).
In Emacs Lisp:
What can be quoted ?
Lists and symbols.
Quoting a number evaluates to the number itself:
'5 is the same as 5.
What happens when you quote lists ?
For example:
'(one two) evaluates to
(list 'one 'two) which evaluates to
(list (intern "one") (intern ("two"))).
(intern "one") creates a symbol named "one" and stores it in a "central" hash-map, so anytime you say 'one then the symbol named "one" will be looked up in that central hash-map.
But what is a symbol ?
For example, in OO-languages (Java/Javascript/Python) a symbol could be represented as an object that has a name field, which is the symbol's name like "one" above, and data and/or code can be associated with it this object.
So an symbol in Python could be implemented as:
class Symbol:
def __init__(self,name,code,value):
self.name=name
self.code=code
self.value=value
In Emacs Lisp for example a symbol can have 1) data associated with it AND (at the same time - for the same symbol) 2) code associated with it - depending on the context, either the data or the code gets called.
For example, in Elisp:
(progn
(fset 'add '+ )
(set 'add 2)
(add add add)
)
evaluates to 4.
Because (add add add) evaluates as:
(add add add)
(+ add add)
(+ 2 add)
(+ 2 2)
4
So, for example, using the Symbol class we defined in Python above, this add ELisp-Symbol could be written in Python as Symbol("add",(lambda x,y: x+y),2).
Many thanks for folks on IRC #emacs for explaining symbols and quotes to me.
Code is data and data is code. There is no clear distinction between them.
This is a classical statement any lisp programmer knows.
When you quote a code, that code will be data.
1 ]=> '(+ 2 3 4)
;Value: (+ 2 3 4)
1 ]=> (+ 2 3 4)
;Value: 9
When you quote a code, the result will be data that represent that code. So, when you want to work with data that represents a program you quote that program. This is also valid for atomic expressions, not only for lists:
1 ]=> 'code
;Value: code
1 ]=> '10
;Value: 10
1 ]=> '"ok"
;Value: "ok"
1 ]=> code
;Unbound variable: code
Supposing you want to create a programming language embedded in lisp -- you will work with programs that are quoted in scheme (like '(+ 2 3)) and that are interpreted as code in the language you create, by giving programs a semantic interpretation. In this case you need to use quote to keep the data, otherwise it will be evaluated in external language.
When we want to pass an argument itself instead of passing the value of the argument then we use quote. It is mostly related to the procedure passing during using lists, pairs and atoms
which are not available in C programming Language ( most people start programming using C programming, Hence we get confused)
This is code in Scheme programming language which is a dialect of lisp and I guess you can understand this code.
(define atom? ; defining a procedure atom?
(lambda (x) ; which as one argument x
(and (not (null? x)) (not(pair? x) )))) ; checks if the argument is atom or not
(atom? '(a b c)) ; since it is a list it is false #f
The last line (atom? 'abc) is passing abc as it is to the procedure to check if abc is an atom or not, but when you pass(atom? abc) then it checks for the value of abc and passses the value to it. Since, we haven't provided any value to it
Quote returns the internal representation of its arguments. After plowing through way too many explanations of what quote doesn't do, that's when the light-bulb went on. If the REPL didn't convert function names to UPPER-CASE when I quoted them, it might not have dawned on me.
So. Ordinary Lisp functions convert their arguments into an internal representation, evaluate the arguments, and apply the function. Quote converts its arguments to an internal representation, and just returns that. Technically it's correct to say that quote says, "don't evaluate", but when I was trying to understand what it did, telling me what it doesn't do was frustrating. My toaster doesn't evaluate Lisp functions either; but that's not how you explain what a toaster does.
Anoter short answer:
quote means without evaluating it, and backquote is quote but leave back doors.
A good referrence:
Emacs Lisp Reference Manual make it very clear
9.3 Quoting
The special form quote returns its single argument, as written, without evaluating it. This provides a way to include constant symbols and lists, which are not self-evaluating objects, in a program. (It is not necessary to quote self-evaluating objects such as numbers, strings, and vectors.)
Special Form: quote object
This special form returns object, without evaluating it.
Because quote is used so often in programs, Lisp provides a convenient read syntax for it. An apostrophe character (‘'’) followed by a Lisp object (in read syntax) expands to a list whose first element is quote, and whose second element is the object. Thus, the read syntax 'x is an abbreviation for (quote x).
Here are some examples of expressions that use quote:
(quote (+ 1 2))
⇒ (+ 1 2)
(quote foo)
⇒ foo
'foo
⇒ foo
''foo
⇒ (quote foo)
'(quote foo)
⇒ (quote foo)
9.4 Backquote
Backquote constructs allow you to quote a list, but selectively evaluate elements of that list. In the simplest case, it is identical to the special form quote (described in the previous section; see Quoting). For example, these two forms yield identical results:
`(a list of (+ 2 3) elements)
⇒ (a list of (+ 2 3) elements)
'(a list of (+ 2 3) elements)
⇒ (a list of (+ 2 3) elements)
The special marker ‘,’ inside of the argument to backquote indicates a value that isn’t constant. The Emacs Lisp evaluator evaluates the argument of ‘,’, and puts the value in the list structure:
`(a list of ,(+ 2 3) elements)
⇒ (a list of 5 elements)
Substitution with ‘,’ is allowed at deeper levels of the list structure also. For example:
`(1 2 (3 ,(+ 4 5)))
⇒ (1 2 (3 9))
You can also splice an evaluated value into the resulting list, using the special marker ‘,#’. The elements of the spliced list become elements at the same level as the other elements of the resulting list. The equivalent code without using ‘`’ is often unreadable. Here are some examples:
(setq some-list '(2 3))
⇒ (2 3)
(cons 1 (append some-list '(4) some-list))
⇒ (1 2 3 4 2 3)
`(1 ,#some-list 4 ,#some-list)
⇒ (1 2 3 4 2 3)
I'm trying to learn and understand the Lisp programming language to a deep level. The function + evaluates its arguments in applicative order:
(+ 1 (+ 1 2))
(+ 1 2) will be evaluated and then (+ 1 3) will be evaluated, but the if function works differently:
(if (> 1 2) (not-defined 1 2) 1)
As the form (not-defined 1 2) isn't evaluated, the program doesn't break.
How can the same syntax lead to different argument evaluation? How is the if function defined so that its arguments aren't evaluated?
if is a special operator, not an ordinary function.
This means that the normal rule that the rest elements in the compound form are evaluated before the function associated with the first element is invoked is not applicable (in that it is similar to macro forms).
The way this is implemented in a compiler and/or an interpreter is that one looks at the compound form and decides what to do with it based on its first element:
if it is a special operator, it does its special thing;
if it is a macro, its macro-function gets the whole form;
otherwise it is treated as a function - even if no function is defined.
Note that some special forms can be defined as macros expanding to other special forms, but some special forms must actually be present.
E.g., one can define if in terms of cond:
(defmacro my-if (condition yes no)
`(cond (,condition ,yes)
(t ,no)))
and vice versa (much more complicated - actually, cond is a macro, usually expanding into a sequence of ifs).
PS. Note that the distinction between system-supplied macros and special operators, while technically crisp and clear (see special-operator-p and macro-function), is ideologically blurred because
An implementation is free to implement a Common Lisp special operator
as a macro. An implementation is free to implement any macro operator
as a special operator, but only if an equivalent definition of the
macro is also provided.
sds's answer answers this question well, but there are a few more general aspects that I think are worth mentioning. As that answer and others have pointed out, if, is built into the language as a special operator, because it really is a kind of primitive. Most importantly, if is not a function.
That said, the functionality of if can be achieved using just functions and normal function calling where all the arguments are evaluated. Thus, conditionals can be implemented in the lambda calculus, on which languages in the family are somewhat based, but which doesn't have a conditional operator.
In the lambda calculus, one can define true and false as functions of two arguments. The arguments are presumed to be functions, and true calls the first of its arguments, and false calls the second. (This is a slight variation of Church booleans which simply return their first or second argument.)
true = λ[x y].(x)
false = λ[x y].(y)
(This is obviously a departure from boolean values in Common Lisp, where nil is false and anything else is true.) The benefit of this, though, is that we can use a boolean value to call one of two functions, depending on whether the boolean is true or false. Consider the Common Lisp form:
(if some-condition
then-part
else-part)
If were were using the booleans as defined above, then evaluating some-condition will produce either true or false, and if we were to call that result with the arguments
(lambda () then-part)
(lambda () else-part)
then only one of those would be called, so only one of then-part and else-part would actually be evaluated. In general, wrapping some forms up in a lambda is a good way to be able delay the evaluation of those forms.
The power of the Common Lisp macro system means that we could actually define an if macro using the types of booleans described above:
(defconstant true
(lambda (x y)
(declare (ignore y))
(funcall x)))
(defconstant false
(lambda (x y)
(declare (ignore x))
(funcall y)))
(defmacro new-if (test then &optional else)
`(funcall ,test
(lambda () ,then)
(lambda () ,else)))
With these definitions, some code like this:
(new-if (member 'a '(1 2 3))
(print "it's a member")
(print "it's not a member"))))
expands to this:
(FUNCALL (MEMBER 'A '(1 2 3)) ; assuming MEMBER were rewritten
(LAMBDA () (PRINT "it's a member")) ; to return `true` or `false`
(LAMBDA () (PRINT "it's not a member")))
In general, if there is some form and some of the arguments aren't getting evaluated, then the (car of the) form is either a Common Lisp special operator or a macro. If you need to write a function where the arguments will be evaluated, but you want some forms not to be evaluated, you can wrap them up in lambda expressions and have your function call those anonymous functions conditionally.
This is a possible way to implement if, if you didn't already have it in the language. Of course, modern computer hardware isn't based on a lambda calculus interpreter, but rather on CPUs that have test and jump instructions, so it's more efficient for the language to provide if a primitive and to compile down to the appropriate machine instructions.
Lisp syntax is regular, much more regular than other languages, but it's still not completely regular: for example in
(let ((x 0))
x)
let is not the name of a function and ((x 0)) is not a bad form in which a list that is not a lambda form has been used in the first position.
There are quite a few "special cases" (still a lot less than other languages, of course) where the general rule of each list being a function call is not followed, and if is one of them. Common Lisp has quite a few "special forms" (because absolute minimality was not the point) but you can get away for example in a scheme dialect with just five of them: if, progn, quote, lambda and set! (or six if you want macros).
While the syntax of Lisp is not totally uniform the underlying representation of code is however quite uniform (just lists and atoms) and the uniformity and simplicity of representation is what facilitates metaprogramming (macros).
"Lisp has no syntax" is a statement with some truth in it, but so it's the statement "Lisp has two syntaxes": one syntax is what uses the reader to convert from character streams to s-expressions, another syntax is what uses the compiler/evaluator to convert from s-expressions to executable code.
It's also true that Lisp has no syntax because neither of those two levels is fixed. Differently from other programming languages you can customize both the first step (using reader macros) and the second step (using macros).
It would not make any sense to do so. Example: (if (ask-user-should-i-quit) (quit) (continue)). Should that quit, even though the user does not want to?
IF is not a function in Lisp. It is a special built-in operator. Lisp a several built-in special operators. See: Special Forms. Those are not functions.
The arguments are not evaluated as for functions, because if is a special operator. Special operators can be evaluated in any arbitrary way, that's why they're called special.
Consider e.g.
(if (not (= x 0))
(/ y x))
If the division was always evaluated, there could be a division by zero error which obviously was not intended.
If isn't a function, it's a special form. If you wanted to implement similar functionality yourself, you could do so by defining a macro rather than a function.
This answer applies to Common Lisp, but it'll probably the same for most other Lisps (though in some if may be a macro rather than a special form).
Why do we have to use funcall to call higher order functions in Common Lisp? For example, why do we have to use:
(defun foo (test-func args)
(funcall test-func args))
instead of the simpler:
(defun bar (test-func args)
(test-func args))
Coming from a procedural background, I'm a bit surprised by that since the languages I'm more used to (e.g. Python, C#) don't need the distinction. In particular, on the source level at least, the C# compiler transforms it to something like func.invoke().
The only problem I see is that this would mean we couldn't call a global function test-func anymore because it'd be shadowed, but that's hardly a problem.
Strictly speaking, funcall would not be needed, but there are some lisps (lisp-2 variants, such as Common Lisp) that separate the variable name space of the function name space. Lisp-1 variants (e.g. Scheme) do not make this distinction.
More specifically, in your case, test-func is in the variable name space.
(defun foo (test-func args)
(funcall test-func args))
Therefore you need a construct that actually searches the function object associated with this variable in the variable name space. In Common Lisp this construct is funcall.
See also this answer.
The majority of Lisps have two namespaces (functions and variables). A name is looked up in the function namespace when it appears as the first element in an S-expression, and in the variable namespace otherwise. This allows you to name your variables without worrying about whether they shadow functions: so you can name your variable list instead of having to mangle it into lst.
However, this means that when you store a function in a variable, you can't call it normally:
(setq list #'+) ; updates list in the variable namespace
(list 1 2 3) => (1 2 3) ; looks up list in the function namespace
Hence the need for funcall and apply:
(funcall list 1 2 3) => 6 ; looks up list in the variable namespace
(Not all Lisps have two namespaces: Scheme is an example of a Lisp with just one namespace.)
In Common Lisp, each symbol can be associated with its symbol-function and its symbol-value, among other things. When reading a list, by default, Common Lisp interprets:
arg1 as a function and so retrieves test-func's symbol-function, which is undefined -- thus function bar doesn't work
arg2 as something to be evaled -- thus function foo retrieves test-func's symbol-value, which, in your case, happens to be a function
Consider this piece of code:
(defvar lst '(1 1))
(defmacro get-x (x lst)
`(nth ,x ,lst))
(defun get-y (y lst)
(nth y lst))
Now let us assume that I want to change the value of the elements of the list called lst, the car with get-x and the cdr with get-y.
As I try to change the value with get-x (with setf) everything goes fine but if I try it with get-y it signals an error (shortened):
; caught STYLE-WARNING:
; undefined function: (SETF GET-STUFF)
Why does this happen?
I myself suspect that this happens because the macro simply expands and the function nth simply returns a reference to the value of an element in the list and the function on the other hand evaluates the function-call to nth and returns the value of the referenced value (sounds confusing).
Am I correct in my suspicions?
If I am correct then how will one know what is simply a reference to a value and an actual value?
The error does not happen with the macro version, because, as you assumed, the expression (setf (get-x some-x some-list) some-value) will be expanded (at compile-time) into something like (setf (nth some-x some-list) some-value) (not really, but the details of setf-expansion are complex), and the compiler knows, how to deal with that (i.e., there is a suitable setf expander defined for function nth).
However, in the case of get-y, the compiler has no setf expander, unless you provide one. The easiest way to do so would be
(defun (setf get-y) (new-value x ls) ; Note the function's name: setf get-y
(setf (nth x ls) new-value))
Note, that there are a few conventions regarding setf-expanders:
The new value is always provided as the first argument to the setf function
All setf functions are supposed to return the new value as their result (as this is, what the entire setf form is supposed to return)
There is, BTW, no such concept as a "reference" in Common Lisp (at least not in the C++ sense), though there once were Lisp dialects which had locatives. Generalized place forms (ie., setf and its machinery) work very differently from plain C++ style references. See the CLHS, if you are curious about the details.
SETF is a macro.
The idea is that to set and read elements from data structures are two operations, but usually require two different names (or maybe even something more complex). SETF now enables you to use just one name for both:
(get-something x)
Above reads a datastructure. The inverse then simply is:
(setf (get-something x) :foobar)
Above sets the datastructure at X with :FOOBAR.
SETF does not treat (get-something x) as a reference or something like that. It just has a database of inverse operations for each operation. If you use GET-SOMETHING, it knows what the inverse operation is.
How does SETF know it? Simple: you have to tell it.
For The NTH operation, SETF knows how to set the nth element. That's builtin into Common Lisp.
For your own GET-Y operation SETF does not have that information. You have to tell it. See the Common Lisp HyperSpec for examples. One example is to use DEFUN and (SETF GET-Y) as a function name.
Also note following style problems with your example:
lst is not a good name for a DEFVAR variable. Use *list* as a name to make clear that it is a special variable declared by DEFVAR (or similar).
'(1 2) is a literal constant. If you write a Common Lisp program, the effects of changing it are undefined. If you want to change a list later, you should cons it with LIST or something like COPY-LIST.