I hope there's someone to help me with this, because I can't find any useful answer, and I'm new with Lisp.
What I'm trying to do is to test a value of one element and to print something if its 1, otherwise to print blank character.
This works when all of the list arguments have the value 1:
(defun print-lst (list)
(format t "~%~a ~a ~a~%"
(if (= (nth 0 list) '1)
'¦)
(if (= (nth 1 list) '1)
'P)
(if (= (nth 2 list) '1)
'¦)))
so the output is ¦ P ¦. But, if the second element in list is 0, it prints NIL on that place ¦ NIL ¦ and I want it to print a space instead ¦ ¦(not just to skip that character¦¦, it is important to there is a blank character in that position in output line if the tested value is not 1).
Is there any way to return a blank character if the condition (if (= (nth 1 list) '1) 'P) is not fulfilled or is there any other way to perform this?
I hope I explained that nicely. Thank you.
If you want to make full use of the power of format, you can use a combination of format conditionals and format GOTO.
Like this:
[1]> (format nil "~#{~:[<nothing>~;~:*~A~]~^ ~}" 1 2 nil 4 nil 6)
"1 2 <nothing> 4 <nothing> 6"
In your case, this should work:
(format t "~&~#{~:[ ~;~:*~A~]~^ ~}"
...)
This works by doing the following:
~& inserts a newline unless we're already at the beginning of a line.
~#{...~} processes the arguments iteratively.
~:[...~;...~] chooses between the nil and non-nil case.
~:* unconsumes the argument that was consumed by ~:[...~].
~A outputs the item being processed.
~^ escapes from the loop on the last iteration (so as not to output an excessive space after the last item).
If takes three arguments: condition, then-form, else-form; the else-form is optional. Besides, I would use literal character syntax for literal characters.
(if (= (nth 0 list) 1)
#\P
#\Space)
Documentation:
Special form if
Character syntax
Character names
Counter question:
Do you really need output values that are identifiers ('something) or would also string literals work ("something")?
If the first is true: I suppose it is not possible to use space as an identifier.
If the second is true: use "|", "P" and " " as output values
Format is a beast waiting to devour the unwary.
That said, it looks like you may want to use some of its higher level directives here. Check out the Formatted Output section of the Lisp Hyperspec, and the format chapter of PCL (specifically, look at the section that deals with conditional formatting).
Related
Writing a simple loop for a Lisp program, I want my numbers to be printed with whitespace in between them,
(loop for i from 1 to 5
do
(prin1 i))
prints:
12345
id like it to print:
1 2 3 4 5
but i cant find any info on printing whitespace anywhere. Any help is appreciated!
All sufficiently-mature programming languages go through a stage of having another programming language embedded in strings, often to control printed output. Really mature programming languages pass this stage: CL is not yet that mature, and still has FORMAT control strings. And you can do all sorts of things with these.
A slight variant of your example is to get FORMAT to print a list:
> (format t "~&~{~A~^, ~}.~%" '(1 2 3 4 5))
1, 2, 3, 4, 5.
What this control string means is:
~& – be at the start of the line;
~{ – start an iteration over a list;
~A – print each element as with PRINC;
~^ – after printing the last element, stop here;
, – print a comma and a space for every other element;
~} – end of the loop over elements;
. – print a stop at the end of the list;
~% – print a newline character.
This is a relatively simple example of using FORMAT control strings: it is possible to do some fairly hairy things.
(In fact, of course, CL being CL it's obviously possible to seamlessly add some more structured output-control language to the language, and I'm sure people have done that. So CL has passed the stage in some sense: FORMAT control strings are just, for some people, a fossilized remmnant).
With a library, printing a string:
(ql:quickload "str")
(str:join " " (loop for i from 1 to 5 collect i))
;; "1 2 3 4 5"
https://github.com/vindarel/cl-str/
more:
https://lispcookbook.github.io/cl-cookbook/strings.html
and a primer on iteration (loop, iterate, for, series, mapcar): https://lispcookbook.github.io/cl-cookbook/iteration.html
but I encourage the use of format.
You will somehow need to print the whitespace yourself. Simple extension of your code:
(loop for i from 1 to 5
do (princ i)
(princ #\space))
(Note that I used princ instead of prin1.)
You can use the format function with format directives:
(loop for i from 1 to 5
do
(format t "~A " i))
your output:
1 2 3 4 5
The ~A directive is simply a place holder for a value, in this case the value stored in the variable i. This is similar to Pythons format function.
CL-USER 67 > (loop with start = 1 and end = 5
for i from (1+ start) to end
initially (princ start)
do (princ #\space) (princ i))
1 2 3 4 5
NIL
So I have been given the task by my tutor to make a small function that returns a description of your zodiac sign but I"m having problems with the final output of the sign description, the output is still in brackets and I don't know how to get them out.
(defparameter *month-signs* '((January Capricorn Aquarius 19 nil)
(February Aquarius Pisces 18 nil)
....)))
(defparameter *sign-traits* '((Capricorn (You are a goat be happy!))
....)))
(defun cookie (month-signs sign-traits)
(format t "Please input your month of birth followed by return, and then your day of birth.
e.g january <ret> 22 <ret> Thank You. ~%")
(let* ((month (read t));reads in month of birth
(day (read t));reads in day of birth
(month-assoc (assoc month month-signs)))
(cond
((>=(cadddr month-assoc)day);checks if your sign is the first part of the month
(format t "You are ~D~%" (cadr month-assoc));if it is, prints your sign and signs description
(format t "Your sign description is: ~D~%" (cadr(assoc(cadr month-assoc) sign-traits))))
((< (cadddr month-assoc)22);same as above but for the other sign
(format t "You are ~D~%" (caddr month-assoc))
(format t "Your sign description is: ~D~%" (cadr(assoc(caddr month-assoc) sign-traits)))))))
It all works dandy except this bit "(cadr(assoc(caddr month-assoc) sign-traits)" which returns what I want but in brackets and all caps.
CL-USER> (cookie *month-signs* *sign-traits*)
Please input your month of birth followed by return, and then your day of birth.
e.g january <ret> 22 <ret> Thank You.
january
12
You are CAPRICORN
Your sign description is: (YOU ARE A GOAT BE HAPPY)
I'm really struggling to work out what I need to get rid of the (YOU ARE A GOAT BE HAPPY) on the last bit, I'd like it to just print "Your sign description is: You are a goat be happy." it's probably something obvious that I've missed :\
One more thing the .... are just for your sake as the variable is large and would take up a large amount of page space, I slimmed it down as they are all laid out the same way.
In the list '(January Capricorn Aquarius 19 nil), the first three elements are symbols, which are printed out by the REPL in UPPERCASE. Symbols are different from strings. When you quote a list like this, the literals are treated as symbols, not strings.
Similarly, '(You are a goat be happy!) is a list of six symbols. It gets printed out as a list (enclosed in parentheses) of uppercase symbols.
If you replace the symbols and lists with strings:
(defparameter *month-signs* '((January "Capricorn" "Aquarius" 19 nil)...
(defparameter *sign-traits* '((Capricorn "You are a goat be happy!")...
you should get the output you want.
read takes the input as a symbol, so you want to leave the association key (January) as a symbol.
Printing a list with symbols
Common Lisp can print symbols in many ways. The function WRITE gives a basic interface to the printer. We don't want to print string quotes and we want to control how a word gets capitalized.
(defun write-sentence (sentence
&aux ; local variable bindings
(*print-escape* nil) ; no escaping
(*print-readably* nil)) ; we don't print for the reader
(write (first sentence) :case :capitalize)
(dolist (word (rest sentence))
(write " ")
(write word :case :downcase))
(write ".")
(values)) ; return no values
CL-USER 94 > (write-sentence '(YOU ARE A GOAT BE HAPPY))
You are a goat be happy.
PRINC
The standard function to print in a human readable form is PRINC. It does not provide options, other than to optionally specify the output stream. Those need to be bound in the printer control variables. Here we need to tell PRINC which case to use:
(defun write-sentence (sentence)
(let ((*print-case* :capitalize))
(princ (first sentence)))
(dolist (word (rest sentence))
(princ " ")
(let ((*print-case* :downcase))
(princ word)))
(princ ".")
(values))
Format
Similar functionality can be written in a more compact way using the function FORMAT, which includes features to iterate over lists. This example is left as an exercise to decipher:
CL-USER 115 > (format t
"~{~#(~A~)~#{ ~(~A~)~}~}."
'(YOU ARE A GOAT BE HAPPY))
You are a goat be happy.
Case in symbols
It's also possible to specify symbols with case. They need to be escaped in the source code:
|This is a single symbol in Common Lisp!|
I had never really thought about whether a symbol could be a number in Lisp, so I played around with it today:
> '1
1
> (+ '1 '1)
2
> (+ '1 1)
2
> (define a '1)
> (+ a 1)
2
The above code is scheme, but it seems to be roughly the same in Common Lisp and Clojure as well. Is there any difference between 1 and quoted 1?
In Common Lisp, '1 is shorthand for (QUOTE 1). When evaluated, (QUOTE something) returns the something part, unevaluated. However, there is no difference between 1 evaluated and 1 unevaluated.
So there is a difference to the reader: '1 reads as (QUOTE 1) and 1 reads as 1. But there is no difference when evaluted.
Numbers are self-evaluating objects. That's why you don't have to worry about quoting them, as you do with, say, lists.
A symbol can be made from any string. If you want the symbol whose name is the single character 1, you can say:
(intern "1")
which prints |1|, suggesting an alternate way to enter it:
'|1|
Quoting prevents expressions from being evaluated until later. For example, the following is not a proper list:
(1 2 3)
This is because Lisp interprets 1 as a function, which it is not. So the list must be quoted:
'(1 2 3)
When you quote a very simple expression such as a number, Lisp effectively does not alter its behavior.
See Wikipedia: Lisp.
Well, they are in fact very different. '1 is however precisely the same as (quote 1). (car ''x) evaluates to the symbol 'quote'.
1 is an S-expression, it's the external representation of a datum, a number 1. To say that 1 is a 'number-object' or an S-expression to enter that object would both be acceptable. Often it is said that 1 is the external representation for the actual number object.
(quote 1) is another S-expression, it's an S-expression for a list whose first element is the symbol 'quote' and whose second element is the number 1. This is where it's already different, syntactic keywords, unlike functions, are not considered objects in the language and they do not evaluate to them.
However, both are external representations of objects (data) which evaluate to the same datum. The number whose external representation is 1, they are however most certainly not the same objects, the same, code, the same datum the same whatever, they just evaluate to the very same thing. Numbers evaluate to themselves. To say that they are the same is to say that:
(+ 1 (* 3 3))
And
(if "Strings are true" (* 5 (- 5 3)) "Strings are not true? This must be a bug!")
Are 'the same', they aren't, they are both different programs which just happen to terminate to the same value, a lisp form is also a program, a form is a datum which is also a program, remember.
Also, I was taught a handy trick once that shows that self-evaluating data are truly not symbols when entered:
(let ((num 4))
(symbol? num) ; ====> evaluates to #f
(symbol? 'num) ; ====> evaluates to #t
(symbol? '4) ; ====> evaluates to #f
(symbol? '#\c) ; #f again, et cetera
(symbol? (car ''x)) ; #t
(symbol? quote) ; error, in most implementations
)
Self evaluating data truly evaluate to themselves, they are not 'predefined symbols' of some sorts.
In Lisp, the apostrophe prevents symbols to be evaluated. Using an apostrophe before a number is not forbidden, it is not necessary as the numbers represent themselves. However, like any other list, it automatically gets transformed to an appropriate function call. The interpreter considers these numbers coincide with their value.
As has been pointed out, there is no difference, as numbers evaluate to themselves. You can confirm this by using eval:
(eval 1) ;=> 1
This is not limited to numbers, by the way. In fact, in Common Lisp, most things evaluate to themselves. It's just that it's very rare for something other than numbers, strings, symbols, and lists to be evaluated. For instance, the following works:
(eval (make-hash-table)) ;equivalent to just (make-hash-table)
In Lisp, quote prevent the following expression to be evaluated. ' is a shorthand for quote. As a result, '1 is same as (quote 1).
However, in Lisp, symbols can never be a number. I mean, 'abc is a symbol, but '123 is not (evaluated into) a symbol. I think this is wrong of the design of Lisp. Another case is not only #t or #f can be used as a Boolean expression.
I'm porting some code from lisp, but I got stuck at this part (apparently that's for mit-scheme)
(define (end-of-sentence? word)
(and (or (char-exist-last? word '#\.)
(char-exist-last? word '#\!)
(char-exist-last? word '#\?))
(not (initial? word))
(or (eq? (peek-char) '#\Space) ;;peek- so test for linefeed isn't affected
(eq? (peek-char) '#\n) ;;note- test for space isn't neccessary
(eq? (read-char) '#\t))))
;;counts the number of sentences in a given file and
;;stops at the given number
;;returns true once at that position, returns error if eof
(define (goto-sentence? file stop)
(define (c-g-iter num)
(cond ((= num stop)
#t)
((end-of-sentence?)
(c-g-iter (+ num 1)))
((not (char-ready?))
(error "EOF reached, number to large in goto-sentence?: " stop))
(else
(c-g-iter num))))
(begin
(open-path file)
(c-g-iter 1)))
Of course I could just skip that and implement what the comments say it does, but just wanted to be sure there's no magic happening in the background. So... how does this function even work -- where is it actually reading the characters? Is it as ugly as I think and does it consume the characters as a side-effect in the last check in end-of-sentence?? Or does the char-ready? actually read something?
But then again - what does (end-of-sentence?) (c-g-iter (+ num 1)) mean, as I don't expect c-g-iter to return a word.
I'm no scheme programmer, but it appears that characters are being consumed in read-char source
end-of-sentence? on the other hand appears to be getting called without a parameter, even though it is declared to take one. I assume that the functions it calls in turn are tolerant to whatever the system provides for unspecified parameters (nil?)
The pair (end-of-sentence?) (c-g-iter (+ num 1)) is a parameter to cond, which you can think like a switch or concise if/else; the first part is a test (end-of-sentence?), and the second is what to execute if true (c-g-iter (+ num 1))
Just adding my voice to the chorus; maybe I can provide some insight.
Some functions that are in these functions are not standard mit-sheme, such as char-exist-last? and initial?.(1) So I can't be sure what they do.
That being said, I think that end-of-sentence? takes in a string (word, so it should be a word) and returns true if its last character is a '!', '? or '.', and the next character after the word is a space, newline or tab character. Also, looking at intial, it probably can't be the first word in the sentence ('A.', for example, shouldn't return true, but 'A dog.' should.)
read-char does indeed 'consume characters' - "Returns the next character available from input-port, updating input-port to point to the following character." (Googled 'read-char mit scheme' to get MIT-Scheme input procedures.)
Per the same source char-ready? works like so: "Returns #t if a character is ready on input-port and returns #f otherwise."
Hope this is at least someone enlightening!
(1) MIT-Scheme Reference
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)