I would like to generate sequential keys that I can use across a number of hash tables. I will call them 'id1','id2' etc. If ht is my hash table then I would like to make symbols from strings as keys. To add an entry to the hash table I want to so something like:
(setf (gethash (make-symbol "id1") ht) 1)
And then access it again with
(gethash 'id1 ht)
I don't think make-symbol is giving me what I want, and the key 'id1' sn't recognised.
What is the best way to make this key?
Error: symbol should be in a package and needs the correct case
In your case we have:
CL-USER 24 > (symbol-name (make-symbol "id0"))
"id0"
CL-USER 25 > (symbol-package (make-symbol "id0"))
NIL
Make sure that you think about the following:
intern the symbol in a package
intern the symbol in the correct package
make sure the symbol has the correct name with the correct case
write symbols with the case you intend to use, possibly you need to escape the symbol to preserve the case
Examples:
uppercased symbol and lowercase symbol name -> not eq
CL-USER 26 > (eq 'id0 (intern "id0" "CL-USER"))
NIL
uppercased symbol and uppercase symbol name -> is eq
CL-USER 27 > (eq 'id0 (intern "ID0" "CL-USER"))
T
an escaped&lowercase symbol and a lowercase symbol name -> is eq
CL-USER 28 > (eq '|id0| (intern "id0" "CL-USER"))
T
make-symbol creates uninterned symbols. It means you will have a unique symbol every time. To get an interned symbol use intern instead.
Related
I am wondering whether or not (string= "abc" nil) is valid in Common Lisp. I noticed that SBCL does not complain even though nil is not a string. (string= '() nil) returns T although both arguments are not strings ...
(SBCL version: 2.2.2)
In Common Lisp the string comparison operators accept “string designators”. According to the Reference Manual, we have:
string designator n. a designator for a string; that is, an object that denotes a string and that is one of: a character (denoting a singleton string that has the character as its only element), a symbol (denoting the string that is its name), or a string (denoting itself).
So the operators accept symbols, and compare their names.
On the other hand, the empty list is equivalent to the symbol NIL:
nil n. the object that is at once the symbol named "NIL" in the COMMON-LISP package, the empty list, the boolean (or generalized boolean) representing false, and the name of the empty type.
So the comparison is equivalent to testing the equality of the strings "NIL" and "NIL", which is obviously true.
So I'm trying to print out a list that looks a little bit something like this (setq lst (list '- '- '- '- '-)) and in the past I used the print command to print out the whole list, however, when printing the whole list there is parenthesis on each side which I do not want to see. I want to use something like (format t) to print every bit of my list and I have something like this set up.
(loop for item from 0 to 4
do (progn
(format t "~X" (nth item lst))
)
)
This code prints out the list perfectly fine like this, ----- but as a mentioned, I want it to print spaces between each element so that it is output like this - - - - -. I used the conditional "~X" because I looked up how to output spaces with the format command and you are apparently supposed to use "~X" but it does not work so if anybody knows how I could put spaces between elements that would be greatly appreciated.
Why not just use the features provided by format:
CL-USER> (defvar *my-list* '(- - - -))
*MY-LIST*
CL-USER> (format nil "~{~A~^ ~}" *my-list*)
"- - - -"
CL-USER> (format t "~{~A~^ ~}" *my-list*)
- - - -
NIL
Here the first call to format outputs to a string to show where the spaces are placed. ~{ is an iteration directive that takes a list as its argument. The directives between the opening ~{ and closing ~} are used repeatedly as a format string for the elements of the input list. The ~^ directive causes an early escape from the iteration context when there are no more arguments; this prevents a trailing space from being added.
The second call to format just outputs to *standard-output*.
Regarding your update, that you posted in the answers to your own post:
First of all, you should edit your post to show us that you found a solution, rather than having us look through all the answers to see how much progress you made on your initial problem.
As it was already mentioned in another answer, you can iterate through the elements of a list using format built-in syntax with ~{~} and ~^ (see the documentation !)
In your own solution, when you iterate over the list using loop, you can put a space at the end of the format string rather than calling format twice ...
You can use loop for <elt> in <list> rather than iterating with the indices, and calling nth at each step - which is slower, and also more verbose.
The loop ... do <stuff> already wraps the <stuff> in what we call an implicit progn, i.e. you do not need to wrap yourself all your instructions in a progn, the loop macro does that for you.
There also exists the macro dolist, which is (arguably) simpler to use in those case when you simply want to iterate over a list.
To be fair, it looks like you are a Common Lisp beginner. In this case, I suggest you read the excellent Practical Common Lisp book, which covers in details the loop macro, the format function, and a lot of basic principles. It is available for free online, and is often recommended to beginners, for good reasons !
Ok I came up with an ingenius solution to my problem which I definitely should've seen before.
(loop for item from 0 to 4
do (progn
(format t "~X" (nth item lst))
(format t " ")
)
)
I didn't realize I could print a space like that but it works perfectly fine. Sorry for wasting you all's time but hopefully someone else can see this if they are having a brain fart like me and thanks to everyone who tried to help.
The following cl-ppcre regular expression generates an error:
(ppcre:scan-to-strings "\[([a-zA-Z0-9_-]+)]" "[has-instance]")
debugger invoked on a CL-PPCRE:PPCRE-SYNTAX-ERROR in thread
#<THREAD "main thread" RUNNING {10010B0523}>:
Expected end of string. at position 16 in string "[([a-zA-Z0-9_-]+)]"
What I was expecting as return values is:
“[has-instance]”
#(“has-instance”)
in order to get at the string within the brackets. Can someone provide a regex correction? Thanks.
The escape character (backslash) only escapes itself and double quotes (§2.4.5 Double-Quote):
If a single escape character is seen, the single escape character is discarded, the next character is accumulated, and accumulation continues.
That means that:
"\[([a-zA-Z0-9_-]+)]"
is parsed the same as the following, where backslash is not present:
"[([a-zA-Z0-9_-]+)]"
The PCRE syntax implemented by CL-PPCRE understands the opening square bracket as a special syntax for character classes, and ends at the next closing bracket.
Thus, the above reads the following as a class:
[([a-zA-Z0-9_-]
The corresponding regex tree is:
CL-USER> (ppcre:parse-string "[([a-zA-Z0-9_-]")
(:CHAR-CLASS #\( #\[ (:RANGE #\a #\z) (:RANGE #\A #\Z) (:RANGE #\0 #\9) #\_ #\-)
Note in particular that the opening parenthesis inside it is treated literally. When the parser encounters the closing parenthesis that follows the above fragment, it interprets it as the end of a register group, but no such group was started, hence the error message at position 16 of the string.
To avoid treating the bracket as a character class, it must be preceded by a literal backslash in the string, as you tried to do, but in order to do so you must write two backslash characters:
CL-USER> (ppcre:parse-string "\\[([a-zA-Z0-9_-]+)]")
(:SEQUENCE #\[
(:REGISTER
(:GREEDY-REPETITION 1 NIL
(:CHAR-CLASS (:RANGE #\a #\z) (:RANGE #\A #\Z) (:RANGE #\0 #\9) #\_ #\-)))
#\])
The closing square brackets needs no backslash.
I encourage you to write regular expressions in Lisp using the tree form, with :regex terms when it improves clarity: it avoids having to deal with the kind of problems that escaping brings. For example:
CL-USER> (ppcre:scan-to-strings
'(:sequence "[" (:register (:regex "[a-zA-Z0-9_-]+")) "]")
"[has-instance]")
"[has-instance]"
#("has-instance")
Double escape the square brackets.
You forgot to (double) escape the closing bracket, too.
(cl-ppcre:scan-to-strings "\\[([a-zA-Z0-9_-]+)\\]" "[has-instance]")
;; "[has-instance]" ;
;; #("has-instance")
For those who are new to common lisp, you import cl-ppcre using quicklisp:
(load "~/quicklisp/setup.list") ;; adjust path to where you installed your quicklisp
(ql:quickload :cl-ppcre)
There's a format directive to zero-pad digits.
cl-user> (format nil "~12,'0d" 27)
"000000000027"
and there's a similar-ish directive to left-align strings while padding them
cl-user> (format nil "~12#<~d~>" 27)
"27 "
Is there a way to do both? That is:
cl-user> (format nil "~12,something,d" 27)
"270000000000"
The naive "~12,'0#<~d~>" does not seem to do what I want here.
cl-user> (format nil "~12,'0#<~d~>" 27)
"27 "
You're close with the last example, but you need some more commas, because tilde less-than takes four arguments, and the pad char is the fourth arguments, but you're passing it as the second. Just pass it as the fourth:
CL-USER> (format nil "~12,,,'0#<~d~>" 27)
"270000000000"
As an aside, it was pointed out in the comments that right padding changes the value that that doesn't seem like a useful operation. I'd say that it can be a useful operation. It might depend on whether these are integers or strings where the values happen to be digit characters. I've seen maintenance systems where upgrades have changed field width and the procedure for migrating old records is to right pad with 0's. The right padding was precisely because it changes the value. 000027 (six chars) can be written as 27, which isn't six chars wide, and 000027 could also be accidentally read (probably by machine, when a programmer isn't careful) as an octal. 270000, on the other hand, has to be six-digits, and won't be octal, since it doesn't start with a 0
Use ~A:
(format nil "~33,,,'0A" 27)
==> "270000000000000000000000000000000"
'(a 'b) gives out the answer (a 'b). How does this work when there is no binding for a (which is unquoted).
This is what happens when we evaluate the expression:
'(a 'b)
=> (a 'b)
The ' quote is shorthand for the quote special form, see the linked documentation for more details:
(quote (a 'b))
=> (a 'b)
As you can see, it prevents the quoted arguments from being evaluated, so it doesn't matter if a is undefined, because a is not interpreted as a variable inside a quoted expression. It's a different thing if we tried to build a list like this:
(list a 'b)
The above will produce an error, because a is an undefined variable, whereas b is a quoted expression. This will work, though - because both elements in the list are quoted:
(list 'a 'b)
=> (a b)
And here's another way to build the list shown in the question:
(list 'a ''b)
=> (a 'b)
The quote that prefixes the list expression prevents the evaluation of the components. If you were to write, say, (list a 'b), you'd end up with a problem if there was no binding.
'something is syntactic sugar for (quote something).
So '(a 'b) is (quote (a (quote b))).
quote is not a function; it is a special form that converts the literal expression to the right into a runtime structure that looks like that expression.
Quoting a number or string literal will evaluate to the same as that literal: (quote 42) evaluates to 42
Quoting an identifier will produce a symbol: (quote apple) evaluates to the same as (string->symbol "apple")
Quoting a parenthesized form will produce a list, whose elements are the result of quoting each of the items in the parentheses: (quote (x 5 (a b))) evaluates as (list (quote x) (quote 5) (quote (a b)))
In (quote (a (quote b))), the second "quote" is not treated specially from any other identifier, because the right side of the first quote is taken literally, and none of it is evaluated as Scheme code.
So, (quote (a (quote b))) evaluates to the same as (list (string->symbol "a") (list (string->symbol "quote") (string->symbol "b"))). In other words, it's a list of 2 elements, the first of which is the symbol "a", and second of which is another list of 2 elements, the first of which is the symbol "quote", and the second of which is the symbol "b".
Then we have to get into how the Scheme REPL prints out values.
When the value is a list, it prints parentheses (( )), with recursively printed result of each of the list items inside.
When the value is a symbol, it prints the string of the symbol.
As a special exception, if it would otherwise print (quote something) (which would happen if it is printing a list of 2 elements, the first of which is the symbol "quote"), it will instead print 'something.
Therefore, the above list, which if not for the exception would be printed as (a (quote b)), will actually be printed as (a 'b).