I know, that, unlike Common lisp, Scheme has one common namespace for variables and functions. But does macroses also fall there?
It could be separated by time, in which they exists. But in compilation time, when macroses are all expanses, there surely are some function, like list, or cons, which hence exists both runtime and compile time.
Then could I for example write the following:
(define (add a b) (+ a b))
(let-syntax ((add (lambda (x)
(syntax-case x ()
((_ a ...) (syntax + a ...))))))
(display (add 1 2 3))
(display (reduce-left add 0 '(1 2 3))))
and get 6 6? Or, vice versa, define the macro, and then define the function? And in such expression: (add 1 2) what happen? Will it be a function call, or macro expansion?
Yes, variables and macros are in the same namespace. (I didn't mention procedures because they are merely values that can stored in variables, like numbers or strings.)
Within the body of your 'let-syntax', 'add' always refers to the macro. Everywhere else in your example, 'add' refers to the procedure.
Note that there are two errors in your code:
The 'syntax' expression is not correct; it should be (syntax (+ a ...)).
In your call to 'reduce-left', it is an error to pass the macro 'add' as an argument to a procedure.
I should mention one complication: if you first define 'add' to be a toplevel procedure, then define some other procedures in terms of 'add', and then later redefine 'add' to be a toplevel macro, the results are not well defined and will vary from one implementation to the next. Similarly if 'add' starts out as a toplevel macro and is later redefined as a procedure.
Related
AND and OR are macros and since macros aren't first class in scheme/racket they cannot be passed as arguments to other functions. A partial solution is to use and-map or or-map. Is it possible to write a function that would take arbitrary macro and turn it into a function so that it can be passed as an argument to another function? Are there any languages that have first class macros?
In general, no. Consider that let is (or could be) implemented as a macro on top of lambda:
(let ((x 1))
(foo x))
could be a macro that expands to
((lambda (x) (foo x)) 1)
Now, what would it look like to convert let to a function? Clearly it is nonsense. What would its inputs be? Its return value?
Many macros will be like this. In fact, any macro that could be routinely turned into a function without losing any functionality is a bad macro! Such a macro should have been a function to begin with.
I agree with #amalloy. If something is written as a macro, it probably does something that functions can't do (e.g., introduce bindings, change evaluation order). So automatically converting arbitrary macro into a function is a really bad idea even if it is possible.
Is it possible to write a function that would take arbitrary macro and turn it into a function so that it can be passed as an argument to another function?
No, but it is somewhat doable to write a macro that would take some macro and turn it into a function.
#lang racket
(require (for-syntax racket/list))
(define-syntax (->proc stx)
(syntax-case stx ()
[(_ mac #:arity arity)
(with-syntax ([(args ...) (generate-temporaries (range (syntax-e #'arity)))])
#'(λ (args ...) (mac args ...)))]))
((->proc and #:arity 2) 42 12)
(apply (->proc and #:arity 2) '(#f 12))
((->proc and #:arity 2) #f (error 'not-short-circuit))
You might also be interested in identifier macro, which allows us to use an identifier as a macro in some context and function in another context. This could be used to create a first class and/or which short-circuits when it's used as a macro, but could be passed as a function value in non-transformer position.
On the topic of first class macro, take a look at https://en.wikipedia.org/wiki/Fexpr. It's known to be a bad idea.
Not in the way you probably expect
To see why, here is a way of thinking about macros: A macro is a function which takes a bit of source code and turns it into another bit of source code: the expansion of the macro. In other words a macro is a function whose domain and range are source code.
Once the source code is fully expanded, then it's fed to either an evaluator or a compiler. Let's assume it's fed to a compiler because it makes the question easier to answer: a compiler itself is simply a function whose domain is source code and whose range is some sequence of instructions for a machine (which may or may not be a real machine) to execute. Those instructions might include things like 'call this function on these arguments'.
So, what you are asking is: can the 'this function' in 'call this function on these arguments' be some kind of macro? Well, yes, it could be, but whatever source code it is going to transform certainly can not be the source code of the program you are executing, because that is gone: all that's left is the sequence of instructions that was the return value of the compiler.
So you might say: OK, let's say we disallow compilers: can we do it now? Well, leaving aside that 'disallowing compilers' is kind of a serious limitation, this was, in fact, something that very old dialects of Lisp sort-of did, using a construct called a FEXPR, as mentioned in another answer. It's important to realise that FEXPRs existed because people had not yet invented macros. Pretty soon, people did invent macros, and although FEXPRs and macros coexisted for a while – mostly because people had written code which used FEXPRs which they wanted to keep running, and because writing macros was a serious pain before things like backquote existed – FEXPRs died out. And they died out because they were semantically horrible: even by the standards of 1960s Lisps they were semantically horrible.
Here's one small example of why FEXPRs are so horrible: Let's say I write this function in a language with FEXPRs:
(define (foo f g x)
(apply f (g x)))
Now: what happens when I call foo? In particular, what happens if f might be a FEXPR?. Well, the answer is that I can't compile foo at all: I have to wait until run-time and make some on-the-fly decision about what to do.
Of course this isn't what these old Lisps with FEXPRs probably did: they would just silently have assumed that f was a normal function (which they would have called an EXPR) and compiled accordingly (and yes, even very old Lisps had compilers). If you passed something which was a FEXPR you just lost: either the thing detected that, or more likely it fall over horribly or gave you some junk answer.
And this kind of horribleness is why macros were invented: macros provide a semantically sane approach to processing Lisp code which allows (eventually, this took a long time to actually happen) minor details like compilation being possible at all, code having reasonable semantics and compiled code having the same semantics as interpreted code. These are features people like in their languages, it turns out.
Incidentally, in both Racket and Common Lisp, macros are explicitly functions. In Racket they are functions which operate on special 'syntax' objects because that's how you get hygiene, but in Common Lisp, which is much less hygienic, they're just functions which operate on CL source code, where the source code is simply made up of lists, symbols &c.
Here's an example of this in Racket:
> (define foo (syntax-rules ()
[(_ x) x]))
> foo
#<procedure:foo>
OK, foo is now just an ordinary function. But it's a function whose domain & range are Racket source code: it expects a syntax object as an argument and returns another one:
> (foo 1)
; ?: bad syntax
; in: 1
; [,bt for context]
This is because 1 is not a syntax object.
> (foo #'(x 1))
#<syntax:readline-input:5:10 1>
> (syntax-e (foo #'(x 1)))
1
And in CL this is even easier to see: Here's a macro definition:
(defmacro foo (form) form)
And now I can get hold of the macro's function and call it on some CL source code:
> (macro-function 'foo)
#<Function foo 4060000B6C>
> (funcall (macro-function 'foo) '(x 1) nil)
1
In both Racket and CL, macros are, in fact, first-class (or, in the case of Racket: almost first-class, I think): they are functions which operate on source code, which itself is first-class: you can write Racket and CL programs which construct and manipulate source code in arbitrary ways: that's what macros are in these languages.
In the case of Racket I have said 'almost first-class', because I can't see a way, in Racket, to retrieve the function which sits behind a macro defined with define-syntax &c.
I've created something like this in Scheme, it's macro that return lambda that use eval to execute the macro:
(define-macro (macron m)
(let ((x (gensym)))
`(lambda (,x)
(eval `(,',m ,#,x)))))
Example usage:
;; normal eval
(define x (map (lambda (x)
(eval `(lambda ,#x)))
'(((x) (display x)) ((y) (+ y y)))))
;; using macron macro
(define x (map (macron lambda)
'(((x) (display x)) ((y) (+ y y)))))
and x in both cases is list of two functions.
another example:
(define-macro (+++ . args)
`(+ ,#args))
((macron +++) '(1 2 3))
Reading this question got me thinking about what constitutes a valid car of an expression. Obviously, symbols and lambdas can be "called" using the usual syntax. According to the hyperspec,
function name n. 1. (in an environment) A symbol or a list (setf symbol) that is the name of a function in that environment. 2. A symbol or a list (setf symbol).
So, theoretically, (setf some-name) is a function name. I decided to give it a try.
(defun (setf try-this) ()
(format t "Don't name your functions like this, kids :)"))
((setf try-this))
(funcall '(setf try-this))
(setf (try-this))
GNU CLISP, SBCL, and ABCL will all let me define this function. However, SBCL and ABCL won't let me call it using any of the syntaxes shown in the snippet. CLISP, on the other hand, will run the first two but still errs on the third.
I'm curious about which compiler is behaving correctly. Since SBCL and ABCL agree, I would hazard a guess that a correct implementation should reject that code. As a second question, how would I call my incredibly-contrived not-useful function from the code snippet, since the things I tried above aren't working portably. Or, perhaps more usefully,
A SETF function has to take at least one argument, which is the new value to be stored in the place. It can take additional arguments as well, these will be filled in from arguments in the place expression in the call to SETF.
When you use SETF, it has to have an even number of arguments: every place you're assigning to needs a value to be assigned.
So it should be:
(defun (setf try-this) (new-value)
(format t "You tried to store ~S~%" new-value))
(setf (try-this) 3)
(funcall #'(setf try-this) 'foo)
You can't use
((setf try-this) 'bar)
because the car of a form does not contain a function name. It can only be a symbol or a lambda expression (although implementations may allow other formats as extensions).
I want to define a list of accumulators with Emacs Lisp and write the following code, but I got a error saying that initV is a void variable. It seems initV is not evaluated in the function define-accum. Where is I make a mistake? (I just want to know why although I know there is other ways to reach my target.)
(defun define-accum (name initV)
(defalias name (lambda (v) (+ v initV))))
(setq accums '((myadd1 . 1)
(myadd2 . 2)))
(dolist (a accums)
(define-accum (car a) (cdr a)))
(message "result = %d" (+ (myadd1 1) (myadd2 1)))
You need to use backquotes properly. This would work for you, for instance:
(defun define-accum (name initV)
(defalias name `(lambda (v) (+ v ,initV))))
See here for an explanation
Apart from using backquotes, you can activate lexical binding (if you're using Emacs 24 or newer). For example, if I put your code in a .el file and put this on the first line:
;; -*- lexical-binding: t -*-
then I get the output:
result = 5
This works because the lambda function in define-accum will reference the initV in the environment where it's being defined (thus picking the variable in the argument list), and create a closure over this variable. With dynamic binding (the default), the function would look for initV in the environment where it's being called.
To add a little to what others have said -
If the variable (initV) is never actually used as a variable, so that in fact its value at the time the accumulator is defined is all that is needed, then there is no need for the lexical closure that encapsulates that variable and its value. In that case, the approach described by #juanleon is sufficient: it uses only the value at definition time - the variable does not exist when the function is invoked (as you discovered).
On the other hand, the lexical-closure approach lets the function be byte-compiled. In the backquote approach, the function is simply represented at runtime by a list that represents a lambda form. If the lambda form represents costly code then it can make sense to use the lexical-closure approach, even though (in this case) the variable is not really needed (as a variable).
But you can always explicitly byte-compile the function (e.g. ##NAME## in your define-accum. That will take care of the inefficiency mentioned in #2, above.
numberp is a predicate in Lisp, and (numberp 1) returns T as expected. But if I only type numberp in the console, it prompts that the variable name is undefined.
What's the difference between these 2?
The question is slightly wrong already.
We are talking about different things in Common Lisp:
symbol : that's a data structure in Lisp. A symbol is a data object with a name, a value, a function, a package and possibly more.
In Common Lisp the symbol can have both a value and a function (or macro).
a variable is an identifier for a value in Lisp code
There are top-level variables defined by DEFVAR and DEFPARAMETER. There are also local variables defined by LAMBDA, DEFUN, LET, LET* and others.
(defun foo (i-am-a-variable) ...)
(defparameter *i-am-a-global-variable* ...)
a named function is an identifier for a function in Lisp code. Named functions are introduced on the top-level by DEFUN, DEFGENERIC and DEFMETHOD. There are also local named functions defined by FLET and LABELS.
Example:
(defun i-am-a-function (foo) ...)
(flet ((i-am-a-function (foo) ...)) ...)
To further complication function names and variable names are symbols in the source code.
Example:
(type-of (second '(defun foo () nil))) --> SYMBOL
Let's look at functions and variables:
(defun foo ()
(let ((some-name 100))
(flet ((some-name (bar) (+ bar 200)))
(+ some-name (some-name some-name)))))
Above code uses a variable and a function with the same symbol in the source code. Since functions and variables have their own namespace, there is no collision.
(+ some-name (some-name some-name))
Above then means we add the variable to the result of the function call on the variable.
This has the practical effect that you can do something like this:
(defun parent (person) ...)
(defun do-something (parent)
(parent parent))
You don't have to fear that your local variable names will shadow a global (or local) function. They simply are in different namespaces.
In Scheme there is only one namespace and we have to write
(define (foo lst) (list 'a lst 'n))
where in Common Lisp we can write:
(defun foo (list) (list 'a list 'n))
In Common Lisp there is no need to write lst instead of list - because there is no clash between the local variable list and the global function list.
Accessing the other namespace
To get a function object stored in a variable you can use FUNCTION.
(let ((my-function (function numberp)))
(funcall my-function 10))
(function foo) can be written shorter as #'foo.
FUNCALL calls a function object.
OTOH, if you want to store a function object from a variable in the function namespace, there is only one way:
(setf (symbol-function 'foo) my-function)
It is also necessary that the object is really a function and not something else (a number, a string, ...). Otherwise you'll see an error.
The side effect of this is that a Common Lisp never has to check if in (foo bar) the FOO is really a function. There is no possibility that it can be other than a function or undefined.
Functions and variables live in different namespaces in Common Lisp.
So when you use a name in a position where a function is expected (i.e. at the head of a list that is being evaluated), it looks for a function (or macro) with that name. If you use the same name at a position where a variable is expected, it looks for a variable with that name.
In your case there is a function named numberp, but not variable named numberp, so the second case causes an error.
One of the differences between Scheme (one Lisp dialect, so to speak) and Common Lisp is that in Scheme, there is one namespace for variables and functions, whereas in CL, there are two separate namespaces.
Thus, in Scheme, "define" sets this one-and-only name-to-value assoc, whereas in CL, there is "define" for the value and "defun" for the function association.
So, in CL you can:
(define foo ...something...)
(defun foo ...somethingElse...)
to confuse the reader.
In Scheme, there is only one:
(define foo something)
If this is good or bad has been an almost religious dispute in the past...
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.