In short: (gethash 'PARIS pandemic-hash-table) is returning nil, despite 'PARIS being a key in the table; this seems to be related to quoting/evaluation of symbols during the hash-table creation in some way but I can't figure it out.
I'm playing with a graph search (testing for the shortest route between cities in the board game Pandemic; just for fun - trying to find optimal research lab placement in a way more sophisticated than 'has the most edges'). I'm using a hash table to hold the route data (nodes and edges), and, needed to input the data as a preliminary:
(defvar *nodes* '('San-Francisco 'Chicago 'Atlanta 'Washington 'Montreal 'New-York 'Madrid 'Paris 'London 'Essen 'Milan 'St-Petersburg))
(defvar *edges* '(('Chicago 'St-Petersburg)
('San-Francisco 'Atlanta 'Montreal)
('Chicago 'Washington)
('Atlanta 'Montreal 'New-York)
('Chicago 'Washington 'New-York)
('Montreal 'Washington 'Madrid 'London)
('New-York 'London 'Paris)
('Madrid 'Essen 'London 'Milan)
('Madrid 'Essen 'London 'New-York)
('London 'Paris 'Milan 'St-Petersburg)
('Paris 'Essen)
('Essen 'Chicago)))
(defvar *pandemic-node-hash* (make-hash-table))
(loop for node in *nodes*
for edges in *edges*
do (setf (gethash node *pandemic-node-hash*) edges))
If I look at the resulting hash table:
CL-USER> (loop for key being the hash-keys of *pandemic-node-hash*
do (print key))
'SAN-FRANCISCO
... ;other keys removed for brevity
'PARIS
NIL
So it's making the table (and the edges show up similarly), but, (gethash 'PARIS *pandemic-node-hash*) returns nil. If I then add another 'PARIS node directly (setf (gethash 'paris *pandemic-node-hash*) 'somevalue), and check the keys, I get:
(loop for key being the hash-keys of *pandemic-node-hash*
do (print key))
'other keys
'PARIS
PARIS
NIL
So, the problem has something to do with the evaluation of the symbols ('PARIS and friends) in the initial hash table creation loop, but I can't quite figure out what's going on or how to do that correctly. I'm guessing node evaluates to the un-evaluated symbol, passing that to gethash ... but what's the right way? Surely not (eval node)? Backtick the list, with commas in front of the symbols? (ugh).
Remember: 'foo is the short form for (quote foo). It's a list with two elements: the symbol CL:QUOTE and the symbol FOO.
(defun show-it (arg)
(print (list arg (type-of arg)))
(values))
Above is a smaller helper function for this answer.
CL-USER 37 > (show-it 'hamburg)
(HAMBURG SYMBOL)
Above shows that the function sees the symbol hamburg and not the value of a variable hamburg.
Not
CL-USER 38 > (show-it '('hamburg))
(((QUOTE HAMBURG)) CONS)
Above: the function sees a nested list, with a list, which has quote as a symbol.
Note: ((quote hamburg)) can be written shorter as('hamburg).
CL-USER 39 > (show-it (first '('hamburg)))
((QUOTE HAMBURG) CONS)
Above: if we get the first element, we get the list with the quote symbol.
Better
CL-USER 40 > (show-it '(hamburg))
((HAMBURG) CONS)
Above provides a list with one symbol, the symbol hamburg.
CL-USER 41 > (show-it (first '(hamburg)))
(HAMBURG SYMBOL)
Above gets the first element, which is the symbol hamburg.
Evaluation
Make sure that you understand evaluation in Lisp. quote blocks evaluation for the whole quoted expression and on all levels of those. Thus it makes no sense to quote contents inside a quoted list.
When passing arguments in a function call, then quoting is used to prevent lists and symbols to be evaluated. Thus quote creates literal data: literal symbols, literal lists, literal vectors, etc.
Thus quoting is a mechanism of code, not of data.
Code:
(first '(hamburg)) vs. (first (hamburg))
getting the first element of a literal list vs.
getting the first element of the result
of calling the function `hamburg`.
Data:
(paris hamburg berlin rome) vs. ('paris 'hamburg 'berlin 'rome)
A list of city names vs. a list of city names,
each nested in a list (quote ...)
Thus:
('paris 'hamburg 'berlin 'rome) makes no sense.
Why not
(dolist (node *nodes*)
(dolist (edges *edges*)
(setf (gethash node *pandemic-node-hash*) edges)))
But also you've double quoted your symbols (as someone else commented)
Why:
'(('Chicago 'St-Petersburg) ...)
when it should probably be just this (don't quote the list and each symbol)
'((Chicago St-Petersburg) ...)
You would see this if you evaluated: *edges*
Related
How does lisp quote work internally?
For example:
(quote (+ 1 (* 1 2)) )
seems to be equivalent to
(list '+ 1 (list '* 1 2))
which means it is some how symbolizing the Head values recursively. Is this function a built in?
Run (equal (quote (+ 1 (* 1 2))) (list '+ 1 (list '* 1 2))) if you don't believe me.
How does it work?
quote is really really simple to implement. It does mostly nothing. The quote special operator just returns the enclosed object like it is. Nothing more. No evaluation. The object is not changed in any way.
Evaluation of quoted forms
Probably a good time to read McCarthy, from 1960:
Recursive Functions of Symbolic Expressions and Their Computation by Machine, Part I
Pages 16/17 explain evaluation with eval. Here:
eq [car [e]; QUOTE] → cadr [e];
or in s-expression notation:
(cond
...
((eq (car e) 'quote)
(cadr e))
...)
Above code implements the evaluation rule for QUOTE: If the expression is a list and the first element of the list is the symbol QUOTE, then return the second element of the list.
Equivalence of a quoted list with a list created by LIST
(equal (quote (+ 1 (* 1 2)))
(list '+ 1 (list '* 1 2)))
The result is T. This means that both result lists are structurally equivalent.
(eq (quote (+ 1 (* 1 2)))
(list '+ 1 (list '* 1 2)))
The result is NIL. This means that the first cons cell of the linked lists are not the same objects. EQ tests whether we really have the same cons cell object.
QUOTE returns a literal data object. The consequences of modifying this object is undefined. So, don't do it.
LIST returns a new freshly consed list each time it is called. The fresh new list will not share any cons cells with any earlier allocated list.
So the main difference is that QUOTE is a built-in operator, which returns literal and unevaluated data. Whereas LIST is a function which creates a new,fresh list with its arguments as contents.
See the effects with respect to EQ and EQUAL:
CL-USER 6 >
(flet ((foo () (quote (+ 1 (* 1 2))))
(bar () (list '+ 1 (list '* 1 2))))
(list (list :eq-foo-foo (eq (foo) (foo)))
(list :eq-foo-bar (eq (foo) (bar)))
(list :eq-bar-bar (eq (foo) (bar)))
(list :equal-foo-foo (equal (foo) (foo)))
(list :equal-foo-bar (equal (foo) (bar)))
(list :equal-bar-bar (equal (foo) (bar)))))
((:EQ-FOO-FOO T)
(:EQ-FOO-BAR NIL)
(:EQ-BAR-BAR NIL)
(:EQUAL-FOO-FOO T)
(:EQUAL-FOO-BAR T)
(:EQUAL-BAR-BAR T))
is quote a function?
quote can't be a function, since it returns its enclosed data unevaluated. Thus it is a special evaluation rule.
If quote were a function, it's arguments were evaluated. But that's exactly what is NOT what quote is supposed to do.
why does Lisp need QUOTE?
Lisp usually uses s-expressions to write Lisp code. So s-expressions have a both purpose to denote data and we use it to write programs. In a Lisp program lists are used for function calls, macro forms and special forms. symbols are used as variables:
(+ n 42)
Here (+ n 42) is a list and n is a symbol. But we also want to use lists as data in our programs and we want to use symbols as data. Thus we have to quote them, so that Lisp will not see them as programs, but as data:
(append '(+ n) '(42)) evaluates to (+ n 42)
Thus in a Lisp program, lists and variables are by default part of the language elements, for example as function calls and variables. If we want to use lists and symbols as literal data, we have to quote them, to prevent the evaluator treating them as Lisp code to evaluate.
quote does nothing more than return its argument unevaluated. But what is an unevaluated argument?
When a Lisp program is defined, it is either read from textual source into s-expression form or constructed directly in terms of s-expressions. A macro would be an example of generating s-expressions. Either way there is a data structure comprising (mostly) symbols and conses that represents the program.
Most Lisp expressions will call upon evaluation and compilation machinery to interpret this data structure as terms in a program. quote is treated specially and passed these uninterpreted symbols and conses as its argument. In short, quote does almost nothing - the value it returns already exists and is simply passed through.
You can observe the difference between passing through and fresh construction by using eq to test the identity of the return value of quote:
(defun f () '(1 2))
(defun g () (list 1 2))
(eq (f) (f)) => T
(eq (g) (g)) => NIL
As you can see, quote returns the same conses each time through.
In Common Lisp, the special operator quote makes whatever followed by un-evaluated, like
(quote a) -> a
(quote {}) -> {}
But why the form (quote ()) gives me nil? I'm using SBCL 1.2.6 and this is what I got in REPL:
CL-USER> (quote ())
NIL
More about this problem: This is some code from PCL Chapter 24
(defun as-keyword (sym)
(intern (string sym) :keyword))
(defun slot->defclass-slot (spec)
(let ((name (first spec)))
`(,name :initarg ,(as-keyword name) :accessor ,name)))
(defmacro define-binary-class (name slots)
`(defclass ,name ()
,(mapcar #'slot->defclass-slot slots)))
When the macro expand for the following code:
(define-binary-class id3-tag
((major-version)))
is
(DEFCLASS ID3-TAG NIL
((MAJOR-VERSION :INITARG :MAJOR-VERSION :ACCESSOR MAJOR-VERSION)))
which is NIL rather than () after the class name ID3-TAG.
nil and () are two ways to express the same concept (the empty list).
Traditionally, nil is used to emphasize the boolean value "false" rather than the empty list, and () is used the other way around.
The Common LISP HyperSpec says:
() ['nil], n. an alternative notation for writing the symbol nil, used
to emphasize the use of nil as an empty list.
Your observation is due to an object to having more than one representation. In Common Lisp the reader (that reads code and reads expressions) parses text to structure and data. When it's data the writer can print it out again but it won't know exactly how the data was represented when it was initially read in. The writer will print one object exactly one way, following defaults and settings, even though there are several representations for that object.
As you noticed nil, NIL, nIL, NiL, ... ,'nil, 'NIL, (), and '() are all read as the very same object. I'm not sure the standard dictates exactly how it's default representation out should be so I guess some implementations choose one of NIL, nil or maybe even ().
With cons the representation is dependent on the cdr being a cons/nil or not:
'(a . nil) ; ==> (a)
'(a . (b . c)) ; ==> (a b . c)
'(a . (b . nil)) ; ==> (a b)
With numbers the reader can get hints about which base you are using. If no base is used in the text it will use whatever *read-base* is:
(let ((*read-base* 2)) ; read numbers as boolean
(read-from-string "(10 #x10)")) ; ==> (2 16)
#x tells the reader to interpret the rest as a hexadecimal value. Now if your print-base would have been 4 the answer to the above would have been visualized as (2 100).
To sum it up.. A single value in Common Lisp may have several good representations and all of them would yield the very same value. How the value is printed will follow both implementation, settings and even arguments to the functions that produce them. Neither what it accepts as values in or the different ways it can visualize the value tells nothing about how the value actually gets stored internally.
I am not sure what is going on here, a macro example in the text.
Basically, not comfortable with how to use get-setf-method, a built-in macro (maybe function?).
To be specific, how about the case that some of the return values of get-setf-method are nil?
e.g.
(get-setf-method 'x)
NIL ;
NIL ;
(#:NEW-3069) ;
(SETQ X #:NEW-3069) ;
X
And why this example code set the fifth return value to the second return value first, for initialization?
Finally how it can handle the order of setting the variables in an expression, such as (aref ar (incf i)
(get-setf-method '(aref ar (incf i)))
(#:G3070 #:G3071) ;
(AR (INCF I)) ;
(#:G3072) ;
(SYSTEM::STORE #:G3070 #:G3071 #:G3072) ;
(AREF #:G3070 #:G3071)
Here is the definition of the macro :
(defmacro sortf (op &rest places)
(let* ((meths (mapcar #'(lambda (p)
(multiple-value-list
(get-setf-method p)))
places))
(temps (apply #'append (mapcar #'third meths))))
`(let* ,(mapcar #'list
(mapcan #'(lambda (m)
(append (first m)
(third m)))
meths)
(mapcan #'(lambda (m)
(append (second m)
(list (fifth m))))
meths))
,#(mapcon #'(lambda (rest)
(mapcar
#'(lambda (arg)
`(unless (,op ,(car rest) ,arg)
(rotatef ,(car rest) ,arg)))
(cdr rest)))
temps)
,#(mapcar #'fourth meths))))
That's actually some older code. get-setf-method was actually replaced by get-setf-expansion as described in Issue SETF-METHOD-VS-SETF-METHOD Writeup. So what you should be interested in these days is get-setf-expansion. The values that it returns are the pieces of code that you need to safely store a value in a location. This is very important because it's very easy to write modyfing macros incorrectly.
As to why some of the values can be nil, one of the examples in the documentation for get-setf-expansion actually shows how some of the values can be nil:
(get-setf-expansion 'x)
;=> NIL, NIL, (#:G0001), (SETQ X #:G0001), X
But what are those values? For that we need to look at the syntax of the documentation:
Syntax:
get-setf-expansion place &optional environment
⇒ vars, vals, store-vars, writer-form, reader-form
Arguments and Values:
place—a place.
environment—an environment object.
vars, vals, store-vars, writer-form, reader-form—a setf expansion.
Those five return values are described in 5.1.1.2 Setf Expansions:
List of temporary variables a list of symbols naming temporary variables to be bound sequentially, as if by let*, to values resulting
from value forms.
List of value forms a list of forms (typically, subforms of the place) which when evaluated yield the values to which the
corresponding temporary variables should be bound.
List of store variables a list of symbols naming temporary store variables which are to hold the new values that will be assigned to
the place.
Storing form a form which can reference both the temporary and the store variables, and which changes the value of the place and
guarantees to return as its values the values of the store variables,
which are the correct values for setf to return.
Accessing form a form which can reference the temporary variables, and which returns the value of the place.
So what do those values in the example mean?
(get-setf-expansion 'x)
;⇒ NIL, NIL, (#:G0001), (SETQ X #:G0001), X
To write to variable x, we don't need any temporary storage, and since there are no temporary values, we don't need any forms to produce values for them. We can notice here that the first and second values are always lists, and they should always have the same length. The third value is a list of store variables. This is a list, because we can actually use setf to modify multiple values, but in this case there's just one. The variables here are where the macro should actually store the new values for the place. Then, it's the writer-form (setq x #:g0001) that will actually take care of putting the value in the place. x, of course, is a simple way of reading the value.
As a more complex example, have a look at this transcript from SBCL:
CL-USER> (defstruct person
person-name)
;⇒ PERSON
CL-USER> (get-setf-expansion '(char (person-name (first (second list-of-list-of-persons))) 3))
; (#:TMP965)
; ((PERSON-NAME (FIRST (SECOND LIST-OF-LIST-OF-PERSONS))))
; (#:NEW964)
; (SB-KERNEL:%CHARSET #:TMP965 3 #:NEW964)
; (CHAR #:TMP965 3)
This means that if we wanted to change the fourth character of the name of the first person in the second list of persons in a list of lists of persons, we could do it with:
(let* ((temp965 (person-name (first (second list-of-list-of-persons))))
(old-char (char tmp965 3))) ; optional
(setq new964 <compute-new-value>)
(sb-kernel:%charset tmp965 3 new964))
We can compute the new value however we want (just fill in for <compute-new-value>), and we can even reference the old value if we want to (by including the optional line). All we need to do is set new964 to the new value, and then execute the writer-form that was given to us.
There are more examples of get-setf-expansion on Stack Overflow:
what is to append as push is to cons, in Lisp?
Writing a destructive macro or function like incf?
Writing a ++ macro in Common Lisp
How could I implement the push macro?
I'm trying to write a Lisp macro that writes a bunch of macros, but I'm having problems generating macro code that uses the splice operator (in build-bind) that expands inside expressions first.
(defmacro define-term-construct (name filter-p list-keywords)
(let* ((do-list-name (output-symbol "do-~a-list" name))
(with-name (output-symbol "with-~a" name))
(do-filter-name (output-symbol "do-~as" name)))
`(progn
(defmacro ,do-list-name
(ls (&key ,#(append list-keywords '(id operation))) &body body)
(with-gensyms (el)
`(loop-list (,el ,ls :id ,id :operation ,operation)
(let (XXX,#(build-bind ,,name ,el))
(when (,',filter-p ,el)
(,',with-name ,el
,#body)))))))))
After the first pass I want to get:
(define-term-construct some some-p (args name))
->
(PROGN
(DEFMACRO DO-SOME-LIST (LS (&KEY ARGS NAME ID OPERATION) &BODY BODY)
(WITH-GENSYMS (EL)
`(LOOP-LIST (,EL ,LS :ID ,ID :OPERATION ,OPERATION)
(LET (,#(BUILD-BIND ,SOME ,EL))
(WHEN (SOME-P ,EL)
(WITH-SOME ,EL
,#BODY)))))))
Any idea what quote/quasiquotes should I use to get the desired code?
The output that you say that you want want to get has unbalanced commas. ,# already balances the backquote, so you cannot have ,SOME and ,EL. That's two levels of unquoting/splicing inside only one level of backquoting.
I suspect you want:
`(WITH-GENSYMS (EL) ... (LET (,#(BUILD-BIND 'SOME EL)) ...))
The some symbol comes in as an argument to the original macro and has to end up as a quoted symbol when passed to the build-bind function. The EL is evaluated straight. It's just a local variable introduced by the WITH-GENSYMS binding construct, and it is not in backquote context anymore because it is inside the splice.
Transliterating that back to the the original outer macro's backquote: SOME becomes ,name:
,#(build-bind ',name el) ;; two commas out balance two backquotes in
The symbol is spliced in under the umbrella of a protecting quote which will make sure it is treated as a symbol and not a variable.
The el does not need to be spliced in; it's not variable material but a hard-coded feature of the template being generated. If you were to put ,el it would look for an el variable in the define-term-construct macro's scope, where no such thing exists.
I want to do a macro in common lisp which is supposed to take in one of its arguments a list made of slots and strings. Here is the prototype :
(defclass time-info ()
((name :initarg name)
(calls :initarg calls)
(second :initarg second)
(consing :initarg consing)
(gc-run-time :initarg gc-run-time)))
(defun print-table (output arg-list time-info-list) ())
The idea is to print a table based on the arg-list which defines its structure. Here is an example of a call to the function:
(print-table *trace-output*
'("|" name "||" calls "|" second "\")
my-time-info-list)
This print a table in ascII on the trace output. The problem, is that I don't know how to explicitely get the elements of the list to use them in the different parts of my macro.
I have no idea how to do this yet, but I'm sure it can be done. Maybe you can help me :)
I would base this on format. The idea is to build a format string
from your arg-list.
I define a helper function for that:
(defun make-format-string-and-args (arg-list)
(let ((symbols ()))
(values (apply #'concatenate 'string
(mapcar (lambda (arg)
(ctypecase arg
(string
(cl-ppcre:regex-replace-all "~" arg "~~"))
(symbol
(push arg symbols)
"~a")))
arg-list))
(nreverse symbols))))
Note that ~ must be doubled in format strings in order to escape them.
The printing macro itself then just produces a mapcar of format:
(defmacro print-table (stream arg-list time-info-list)
(let ((time-info (gensym)))
(multiple-value-bind (format-string arguments)
(make-format-string-and-args arg-list)
`(mapcar (lambda (,time-info)
(format ,stream ,format-string
,#(mapcar (lambda (arg)
(list arg time-info))
arguments)))
,time-info-list)))
You can then call it like this:
(print-table *trace-output*
("|" name "||" calls "|" second "\\")
my-time-info-list)
Please note the following errors in your code:
You need to escape \ in strings.
Second is already a function name exported from the common-lisp
package. You should not clobber that with a generic function.
You need to be more precise with your requirements. Macros and Functions are different things. Arrays and Lists are also different.
We need to iterate over the TIME-INFO-LIST. So that's the first DOLIST.
The table has a description for a line. Each item in the description is either a slot-name or a string. So we iterate over the description. That's the second DOLIST. A string is just printed. A symbol is a slot-name, where we retrieve the slot-value from the current time-info instance.
(defun print-table (stream line-format-description time-info-list)
(dolist (time-info time-info-list)
(terpri stream)
(dolist (slot-or-string line-format-description)
(princ (etypecase slot-or-string
(string slot-or-string)
(symbol (slot-value time-info slot-or-string)))
stream))))
Test:
> (print-table *standard-output*
'("|" name "||" calls "|" second "\\")
(list (make-instance 'time-info
:name "foo"
:calls 100
:second 10)
(make-instance 'time-info
:name "bar"
:calls 20
:second 20)))
|foo||100|10\
|bar||20|20\
First, you probably don't want the quote there, if you're using a macro (you do want it there if you're using a function, however). Second, do you want any padding between your separators and your values? Third, you're probably better off with a function, rather than a macro.
You also seem to be using "array" and "list" interchangeably. They're quite different things in Common Lisp. There are operations that work on generic sequences, but typically you would use one way of iterating over a list and another to iterate over an array.