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FAQ: Lisp Frequently Asked Questions 3/7 [Monthly posting]

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Archive-name: lisp-faq/part3
Last-Modified: Mon May 22 09:42:48 1995 by Mark Kantrowitz
Version: 1.52
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See reader questions & answers on this topic! - Help others by sharing your knowledge
;;; ****************************************************************
;;; Answers to Frequently Asked Questions about Lisp ***************
;;; ****************************************************************
;;; Written by Mark Kantrowitz and Barry Margolin
;;; lisp_3.faq

This post contains Part 3 of the Lisp FAQ.

If you think of questions that are appropriate for this FAQ, or would
like to improve an answer, please send email to us at

This section contains a list of common pitfalls. Pitfalls are aspects
of Common Lisp which are non-obvious to new programmers and often
seasoned programmers as well.

Common Pitfalls (Part 3):

  [3-0]  Why does (READ-FROM-STRING "foobar" :START 3) return FOOBAR
         instead of BAR?  
  [3-1]  Why can't it deduce from (READ-FROM-STRING "foobar" :START 3)
         that the intent is to specify the START keyword parameter
         rather than the EOF-ERROR-P and EOF-VALUE optional parameters?   
  [3-2]  Why can't I apply #'AND and #'OR?
  [3-3]  I used a destructive function (e.g. DELETE, SORT), but it
         didn't seem to work.  Why? 
  [3-4]  After I NREVERSE a list, it's only one element long.  After I
         SORT a list, it's missing things.  What happened? 
  [3-5]  Why does (READ-LINE) return "" immediately instead of waiting
         for me to type a line?  
  [3-6]  I typed a form to the read-eval-print loop, but nothing happened. Why?
  [3-7]  DEFMACRO doesn't seem to work.
         When I compile my file, LISP warns me that my macros are undefined
         functions, or complains "Attempt to call <function> which is 
         defined as a macro.
  [3-8]  Name conflict errors are driving me crazy! (EXPORT, packages)
  [3-9]  Closures don't seem to work properly when referring to the
         iteration variable in DOLIST, DOTIMES, DO and LOOP.
  [3-10] What is the difference between FUNCALL and APPLY?
  [3-11] Miscellaneous things to consider when debugging code.
  [3-12] When is it right to use EVAL?
  [3-13] Why does my program's behavior change each time I use it?
  [3-14] When producing formatted output in Lisp, where should you put the
         newlines (e.g., before or after the line, FRESH-LINE vs TERPRI,
         ~& vs ~% in FORMAT)?
  [3-15] I'm using DO to do some iteration, but it doesn't terminate. 
  [3-16] My program works when interpreted but not when compiled!

Search for \[#\] to get to question number # quickly.

Subject: [3-0] Why does (READ-FROM-STRING "foobar" :START 3) return FOOBAR instead of BAR? READ-FROM-STRING is one of the rare functions that takes both &OPTIONAL and &KEY arguments: READ-FROM-STRING string &OPTIONAL eof-error-p eof-value &KEY :start :end :preserve-whitespace When a function takes both types of arguments, all the optional arguments must be specified explicitly before any of the keyword arguments may be specified. In the example above, :START becomes the value of the optional EOF-ERROR-P parameter and 3 is the value of the optional EOF-VALUE parameter. To get the desired result, you should use (READ-FROM-STRING "foobar" t nil :START 3) If you need to understand and use the optional arguments, please refer to CLTL2 under READ-FROM-STRING, otherwise, this will behave as desired for most purposes.
Subject: [3-1] Why can't it deduce from (READ-FROM-STRING "foobar" :START 3) that the intent is to specify the START keyword parameter rather than the EOF-ERROR-P and EOF-VALUE optional parameters? In Common Lisp, keyword symbols are first-class data objects. Therefore, they are perfectly valid values for optional parameters to functions. There are only four functions in Common Lisp that have both optional and keyword parameters (they are PARSE-NAMESTRING, READ-FROM-STRING, WRITE-LINE, and WRITE-STRING), so it's probably not worth adding a nonorthogonal kludge to the language just to make these functions slightly less confusing; unfortunately, it's also not worth an incompatible change to the language to redefine those functions to use only keyword arguments.
Subject: [3-2] Why can't I apply #'AND and #'OR? Here's the simple, but not necessarily satisfying, answer: AND and OR are macros, not functions; APPLY and FUNCALL can only be used to invoke functions, not macros and special operators. OK, so what's the *real* reason? The reason that AND and OR are macros rather than functions is because they implement control structure in addition to computing a boolean value. They evaluate their subforms sequentially from left/top to right/bottom, and stop evaluating subforms as soon as the result can be determined (in the case of AND, as soon as a subform returns NIL; in the case of OR, as soon as one returns non-NIL); this is referred to as "short circuiting" in computer language parlance. APPLY and FUNCALL, however, are ordinary functions; therefore, their arguments are evaluated automatically, before they are called. Thus, were APPLY able to be used with #'AND, the short-circuiting would be defeated. Perhaps you don't really care about the short-circuiting, and simply want the functional, boolean interpretation. While this may be a reasonable interpretation of trying to apply AND or OR, it doesn't generalize to other macros well, so there's no obvious way to have the Lisp system "do the right thing" when trying to apply macros. The only function associated with a macro is its expander function; this function accepts and returns and form, so it cannot be used to compute the value. The Common Lisp functions EVERY and SOME can be used to get the functionality you intend when trying to apply #'AND and #'OR. For instance, the erroneous form: (apply #'and *list*) can be translated to the correct form: (every #'identity *list*)
Subject: [3-3] I used a destructive function (e.g. DELETE, SORT), but it didn't seem to work. Why? I assume you mean that it didn't seem to modify the original list. There are several possible reasons for this. First, many destructive functions are not *required* to modify their input argument, merely *allowed* to; in some cases, the implementation may determine that it is more efficient to construct a new result than to modify the original (this may happen in Lisp systems that use "CDR coding", where RPLACD may have to turn a CDR-NEXT or CDR-NIL cell into a CDR-NORMAL cell), or the implementor may simply not have gotten around to implementing the destructive version in a truly destructive manner. Another possibility is that the nature of the change that was made involves removing elements from the front of a list; in this case, the function can simply return the appropriate tail of the list, without actually modifying the list. And example of this is: (setq *a* (list 3 2 1)) (delete 3 *a*) => (2 1) *a* => (3 2 1) Similarly, when one sorts a list, SORT may destructively rearrange the pointers (cons cells) that make up the list. SORT then returns the cons cell that now heads the list; the original cons cell could be anywhere in the list. The value of any variable that contained the original head of the list hasn't changed, but the contents of that cons cell have changed because SORT is a destructive function: (setq *a* (list 2 1 3)) (sort *a* #'<) => (1 2 3) *a* => (2 3) In both cases, the remedy is the same: store the result of the function back into the place whence the original value came, e.g. (setq *a* (delete 3 *a*)) *a* => (2 1) Why don't the destructive functions do this automatically? Recall that they are just ordinary functions, and all Lisp functions are called by value. They see the value of the argument, not the argument itself. Therefore, these functions do not know where the lists they are given came from; they are simply passed the cons cell that represents the head of the list. Their only obligation is to return the new cons cell that represents the head of the list. Thus "destructive" just means that the function may munge the list by modifying the pointers in the cars and cdrs of the list's cons cells. This can be more efficient, if one doesn't care whether the original list gets trashed or not. One thing to be careful about when doing this (storing the result back into the original location) is that the original list might be referenced from multiple places, and all of these places may need to be updated. For instance: (setq *a* (list 3 2 1)) (setq *b* *a*) (setq *a* (delete 3 *a*)) *a* => (2 1) *b* => (3 2 1) ; *B* doesn't "see" the change (setq *a* (delete 1 *a*)) *a* => (2) *b* => (3 2) ; *B* sees the change this time, though One may argue that destructive functions could do what you expect by rearranging the CARs of the list, shifting things up if the first element is being deleted, as they are likely to do if the argument is a vector rather than a list. In many cases they could do this, although it would clearly be slower. However, there is one case where this is not possible: when the argument or value is NIL, and the value or argument, respectively, is not. It's not possible to transform the object referenced from the original cell from one data type to another, so the result must be stored back. Here are some examples: (setq *a* (list 3 2 1)) (delete-if #'numberp *a*) => NIL *a* => (3 2 1) (setq *a* nil *b* '(1 2 3)) (nconc *a* *b*) => (1 2 3) *a* => NIL The names of most destructive functions (except for sort, delete, rplaca, rplacd, and setf of accessor functions) have the prefix N. In summary, the two common problems to watch out for when using destructive functions are: 1. Forgetting to store the result back. Even though the list is modified in place, it is still necessary to store the result of the function back into the original location, e.g., (setq foo (delete 'x foo)) If the original list was stored in multiple places, you may need to store it back in all of them, e.g. (setq bar foo) ... (setq foo (delete 'x foo)) (setq bar foo) 2. Sharing structure that gets modified. If it is important to preserve the shared structure, then you should either use a nondestructive operation or copy the structure first using COPY-LIST or COPY-TREE. (setq bar (cdr foo)) ... (setq foo (sort foo #'<)) ;;; now it's not safe to use BAR Note that even nondestructive functions, such as REMOVE, and UNION, can return a result which shares structure with an argument. Nondestructive functions don't necessarily copy their arguments; they just don't modify them.
Subject: [3-4] After I NREVERSE a list, it's only one element long. After I SORT a list, it's missing things. What happened? These are particular cases of the previous question. Many NREVERSE and SORT implementations operate by rechaining all the CDR links in the list's backbone, rather than by replacing the CARs. In the case of NREVERSE, this means that the cons cell that was originally first in the list becomes the last one. As in the last question, the solution is to store the result back into the original location.
Subject: [3-5] Why does (READ-LINE) return "" immediately instead of waiting for me to type a line? Many Lisp implementations on line-buffered systems do not discard the newline that the user must type after the last right parenthesis in order for the line to be transmitted from the OS to Lisp. Lisp's READ function returns immediately after seeing the matching ")" in the stream. When READLINE is called, it sees the next character in the stream, which is a newline, so it returns an empty line. If you were to type "(read-line)This is a test" the result would be "This is a test". The simplest solution is to use (PROGN (CLEAR-INPUT) (READ-LINE)). This discards the buffered newline before reading the input. However, it would also discard any other buffered input, as in the "This is a test" example above; some implementation also flush the OS's input buffers, so typeahead might be thrown away.
Subject: [3-6] I typed a form to the read-eval-print loop, but nothing happened. Why? There's not much to go on here, but a common reason is that you haven't actually typed a complete form. You may have typed a doublequote, vertical bar, "#|" comment beginning, or left parenthesis that you never matched with another doublequote, vertical bar, "|#", or right parenthesis, respectively. Try typing a few right parentheses followed by Return.
Subject: [3-7] DEFMACRO doesn't seem to work. When I compile my file, LISP warns me that my macros are undefined functions, or complains "Attempt to call <function> which is defined as a macro." When you evaluate a DEFMACRO form or proclaim a function INLINE, it doesn't go back and update code that was compiled under the old definition. When redefining a macro, be sure to recompile any functions that use the macro. Also be sure that the macros used in a file are defined before any forms in the same file that use them. Certain forms, including LOAD, SET-MACRO-CHARACTER, and REQUIRE, are not normally evaluated at compile time. Common Lisp requires that macros defined in a file be used when compiling later forms in the file. If a Lisp doesn't follow the standard, it may be necessary to wrap an EVAL-WHEN form around the macro definition. Most often the "macro was previously called as a function" problem occurs when files were compiled/loaded in the wrong order. For example, developers may add the definition to one file, but use it in a file which is compiled/loaded before the definition. To work around this problem, one can either fix the modularization of the system, or manually recompile the files containing the forward references to macros. Also, if your macro calls functions at macroexpand time, those functions may need to be in an EVAL-WHEN. For example, (defun some-function (x) x) (defmacro some-macro (y) (let ((z (some-function y))) `(print ',z))) If the macros are defined in a file you require, make sure your require or load statement is in an appropriate EVAL-WHEN. Many people avoid all this nonsense by making sure to load all their files before compiling them, or use a system facility (or just a script file) that loads each file before compiling the next file in the system.
Subject: [3-8] Name conflict errors are driving me crazy! (EXPORT, packages) If a package tries to export a symbol that's already defined, it will report an error. You probably tried to use a function only to discover that you'd forgotten to load its file. The failed attempt at using the function caused its symbol to be interned. So now, when you try to load the file, you get a conflict. Unfortunately, understanding and correcting the code which caused the export problem doesn't make those nasty error messages go away. That symbol is still interned where it shouldn't be. Use unintern to remove the symbol from a package before reloading the file. Also, when giving arguments to REQUIRE or package functions, use strings or keywords, not symbols: (find-package "FOO"), (find-package :foo). A sometimes useful technique is to rename (or delete) a package that is "too messed up". Then you can reload the relevant files into a "clean" package.
Subject: [3-9] Closures don't seem to work properly when referring to the iteration variable in DOLIST, DOTIMES, DO and LOOP. DOTIMES, DOLIST, DO and LOOP all use assignment instead of binding to update the value of the iteration variables. So something like (let ((l nil)) (dotimes (n 10) (push #'(lambda () n) l))) will produce 10 closures over the same value of the variable N. To avoid this problem, you'll need to create a new binding after each assignment: (let ((l nil)) (dotimes (n 10) (let ((n n)) (push #'(lambda () n) l)))) Then each closure will be over a new binding of n. This is one reason why programmers who use closures prefer MAPC and MAPCAR to DOLIST.
Subject: [3-10] What is the difference between FUNCALL and APPLY? FUNCALL is useful when the programmer knows the length of the argument list, but the function to call is either computed or provided as a parameter. For instance, a simple implementation of MEMBER-IF (with none of the fancy options) could be written as: (defun member-if (predicate list) (do ((tail list (cdr tail))) ((null tail)) (when (funcall predicate (car tail)) (return-from member-if tail)))) The programmer is invoking a caller-supplied function with a known argument list. APPLY is needed when the argument list itself is supplied or computed. Its last argument must be a list, and the elements of this list become individual arguments to the function. This frequently occurs when a function takes keyword options that will be passed on to some other function, perhaps with application-specific defaults inserted. For instance: (defun open-for-output (pathname &rest open-options) (apply #'open pathname :direction :output open-options)) FUNCALL could actually have been defined using APPLY: (defun funcall (function &rest arguments) (apply function arguments))
Subject: [3-11] Miscellaneous things to consider when debugging code. This question lists a variety of problems to watch out for when debugging code. This is sort of a catch-all question for problems too small to merit a question of their own. See also question [1-3] for some other common problems. Functions: * (flet ((f ...)) (eq #'f #'f)) can return false. * The function LIST-LENGTH is not a faster, list-specific version of the sequence function LENGTH. It is list-specific, but it's slower than LENGTH because it can handle circular lists. * Don't confuse the use of LISTP and CONSP. CONSP tests for the presence of a cons cell, but will return NIL when called on NIL. LISTP could be defined as (defun listp (x) (or (null x) (consp x))). * Use the right test for equality: EQ tests if the objects are identical -- numbers with the same value need not be EQ, nor are two similar lists necessarily EQ. Similarly for characters and strings. For instance, (let ((x 1)) (eq x x)) is not guaranteed to return T. EQL Like EQ, but is also true if the arguments are numbers of the same type with the same value or character objects representing the same character. (eql -0.0 0.0) is not guaranteed to return T. EQUAL Tests if the arguments are structurally isomorphic, using EQUAL to compare components that are conses, bit-vectors, strings or pathnames, and EQ for all other data objects (except for numbers and characters, which are compared using EQL). Except for strings and bit-vectors, arrays are EQUAL only if they are EQ. EQUALP Like EQUAL, but ignores type differences when comparing numbers and case differences when comparing characters. = Compares the values of two numbers even if they are of different types. CHAR= Case-sensitive comparison of characters. CHAR-EQUAL Case-insensitive comparison of characters. STRING= Compares two strings, checking if they are identical. It is case sensitive. STRING-EQUAL Like STRING=, but case-insensitive. * Some destructive functions that you think would modify CDRs might modify CARs instead. (E.g., NREVERSE.) * READ-FROM-STRING has some optional arguments before the keyword parameters. If you want to supply some keyword arguments, you have to give all of the optional ones too. * If you use the function READ-FROM-STRING, you should probably bind *READ-EVAL* to NIL. Otherwise an unscrupulous user could cause a lot of damage by entering #.(shell "cd; rm -R *") at a prompt. * Only functional objects can be funcalled in CLtL2, so a lambda expression '(lambda (..) ..) is no longer suitable. Use #'(lambda (..) ..) instead. If you must use '(lambda (..) ..), coerce it to type FUNCTION first using COERCE. Methods: * PRINT-OBJECT methods can make good code look buggy. If there is a problem with the PRINT-OBJECT methods for one of your classes, it could make it seem as though there were a problem with the object. It can be very annoying to go chasing through your code looking for the cause of the wrong value, when the culprit is just a bad PRINT-OBJECT method. Initialization: * Don't count on array elements being initialized to NIL, if you don't specify an :initial-element argument to MAKE-ARRAY. For example, (make-array 10) => #(0 0 0 0 0 0 0 0 0 0) Iteration vs closures: * DO and DO* update the iteration variables by assignment; DOLIST and DOTIMES are allowed to use assignment (rather than a new binding). (All CLtL1 says of DOLIST and DOTIMES is that the variable "is bound" which has been taken as _not_ implying that there will be separate bindings for each iteration.) Consequently, if you make closures over an iteration variable in separate iterations they may nonetheless be closures over the same variable and hence will all refer to the same value -- whatever value the variable was given last. For example, (let ((fns '())) (do ((x '(1 2) (cdr x))) ((null x)) (push #'(lambda () x) fns)) (mapcar #'funcall (reverse fns))) returns (nil nil), not (1 2), not even (2 2). Thus (let ((l nil)) (dolist (a '(1 2 3) l) (push #'(lambda () a) l))) returns a list of three closures closed over the same bindings, whereas (mapcar #'(lambda (a) #'(lambda () a)) '(1 2 3)) returns a list of closures over distinct bindings. Defining Variables and Constants: * (defvar var init) assigns to the variable only if it does not already have a value. So if you edit a DEFVAR in a file and reload the file only to find that the value has not changed, this is the reason. (Use DEFPARAMETER if you want the value to change upon reloading.) DEFVAR is used to declare a variable that is changed by the program; DEFPARAMETER is used to declare a variable that is normally constant, but which can be changed to change the functioning of a program. * DEFCONSTANT has several potentially unexpected properties: - Once a name has been declared constant, it cannot be used a the name of a local variable (lexical or special) or function parameter. Really. See page 87 of CLtL2. - A DEFCONSTANT cannot be re-evaluated (eg, by reloading the file in which it appears) unless the new value is EQL to the old one. Strictly speaking, even that may not be allowed. (DEFCONSTANT is "like DEFPARAMETER" and hence does an assignment, which is not allowed if the name has already been declared constant by DEFCONSTANT.) Note that this makes it difficult to use anything other than numbers, symbols, and characters as constants. - When compiling (DEFCONSTANT name form) in a file, the form may be evaluated at compile-time, load-time, or both. (You might think it would be evaluated at compile-time and the _value_ used to obtain the object at load-time, but it doesn't have to work that way.) Declarations: * You often have to declare the result type to get the most efficient arithmetic. Eg, (the fixnum (+ (the fixnum e1) (the fixnum e2))) rather than (+ (the fixnum e1) (the fixnum e2)) * Declaring the iteration variable of a DOTIMES to have type FIXNUM does not guarantee that fixnum arithmetic will be used. That is, implementations that use fixnum-specific arithmetic in the presence of appropriate declaration may not think _this_ declaration is sufficient. It may help to declare that the limit is also a fixnum, or you may have to write out the loop as a DO and add appropriate declarations for each operation involved. FORMAT related errors: * When printing messages about files, filenames like foo~ (a GNU-Emacs backup file) may cause problems with poorly coded FORMAT control strings. * Beware of using an ordinary string as the format string, i.e., (format t string), rather than (format t "~A" string). * FORMAT returns NIL, so if you added a format statement at the end of a function for debugging purposes, and that function normally returns a value to the caller, you may have changed the behavior of your program. Miscellaneous: * Be careful of circular lists and shared list structure. * Watch out for macro redefinitions. * A NOTINLINE may be needed if you want SETF of SYMBOL-FUNCTION to affect calls within a file. (See CLtL2, page 686.) * When dividing two numbers, beware of creating a rational number where you intended to get an integer or floating point number. Use TRUNCATE or ROUND to get an integer and FLOAT to ensure a floating point number. This is a major source of errors when porting ZetaLisp or C code to Common Lisp. * If your code doesn't work because all the symbols are mysteriously in the keyword package, one of your comments has a colon (:) in it instead of a semicolon (;). * If you redefine a function while in the debugger, the redefinition may not take effect immediately. This will happen, for example, when the execution stack is halted near the invocation of the function. The function pointer on the stack will still be pointing to the old definition. Go up the stack a few levels before restarting to avoid reusing the old definition.
Subject: [3-12] When is it right to use EVAL? Hardly ever. Any time you think you need to use EVAL, think hard about it. EVAL is useful when implementing a facility that provides an external interface to the Lisp interpreter. For instance, many Lisp-based editors provide a command that prompts for a form and displays its value. Inexperienced macro writers often assume that they must explicitly EVAL the subforms that are supposed to be evaluated, but this is not so; the correct way to write such a macro is to have it expand into another form that has these subforms in places that will be evaluated by the normal evaluation rules. Explicit use of EVAL in a macro is likely to result in one of two problems: the dreaded "double evaluation" problem, which may not show up during testing if the values of the expressions are self-evaluating constants (such as numbers); or evaluation at compile time rather than runtime. For instance, if Lisp didn't have IF and one desired to write it, the following would be wrong: (defmacro if (test then-form &optional else-form) ;; this evaluates all the subforms at compile time, and at runtime ;; evaluates the results again. `(cond (,(eval test) ,(eval then-form)) (t ,(eval else-form)))) (defmacro if (test then-form &optional else-form) ;; this double-evaluates at run time `(cond ((eval ,test) (eval ,then-form)) (t (eval ,else-form))) This is correct: (defmacro if (test then-form &optional else-form) `(cond (,test ,then-form) (t ,else-form))) The following question (taken from an actual post) is typical of the kind of question asked by a programmer who is misusing EVAL: I would like to be able to quote all the atoms except the first in a list of atoms. The purpose is to allow a function to be read in and evaluated as if its arguments had been quoted. This is the wrong approach to solving the problem. Instead, he should APPLY the CAR of the form to the CDR of the form. Then quoting the rest of the form is unnecessary. But one wonders why he's trying to solve this problem in the first place, since the toplevel REP loop already involves a call to EVAL. One gets the feeling that if we knew more about what he's trying to accomplish, we'd be able to point out a more appropriate solution that uses neither EVAL nor APPLY. On the other hand, EVAL can sometimes be necessary when the only portable interface to an operation is a macro.
Subject: [3-13] Why does my program's behavior change each time I use it? Most likely your program is altering itself, and the most common way this may happen is by performing destructive operations on embedded constant data structures. For instance, consider the following: (defun one-to-ten-except (n) (delete n '(1 2 3 4 5 6 7 8 9 10))) (one-to-ten-except 3) => (1 2 4 5 6 7 8 9 10) (one-to-ten-except 5) => (1 2 4 6 7 8 9 10) ; 3 is missing The basic problem is that QUOTE returns its argument, *not* a copy of it. The list is actually a part of the lambda expression that is in ONE-TO-TEN-EXCEPT's function cell, and any modifications to it (e.g., by DELETE) are modifications to the actual object in the function definition. The next time that the function is called, this modified list is used. In some implementations calling ONE-TO-TEN-EXCEPT may even result in the signalling of an error or the complete aborting of the Lisp process. Some Lisp implementations put self-evaluating and quoted constants onto memory pages that are marked read-only, in order to catch bugs such as this. Details of this behavior may vary even within an implementation, depending on whether the code is interpreted or compiled (perhaps due to inlined DEFCONSTANT objects or constant folding optimizations). All of these behaviors are allowed by the draft ANSI Common Lisp specification, which specifically states that the consequences of modifying a constant are undefined (X3J13 vote CONSTANT-MODIFICATION:DISALLOW). To avoid these problems, use LIST to introduce a list, not QUOTE. QUOTE should be used only when the list is intended to be a constant which will not be modified. If QUOTE is used to introduce a list which will later be modified, use COPY-LIST to provide a fresh copy. For example, the following should all work correctly: o (remove 4 (list 1 2 3 4 1 3 4 5)) o (remove 4 '(1 2 3 4 1 3 4 5)) ;; Remove is non-destructive. o (delete 4 (list 1 2 3 4 1 3 4 5)) o (let ((x (list 1 2 4 1 3 4 5))) (delete 4 x)) o (defvar *foo* '(1 2 3 4 1 3 4 5)) (delete 4 (copy-list *foo*)) (remove 4 *foo*) (let ((x (copy-list *foo*))) (delete 4 x)) The following, however, may not work as expected: o (delete 4 '(1 2 3 4 1 3 4 5)) Note that similar issues may also apply to hard-coded strings. If you want to modify elements of a string, create the string with MAKE-STRING.
Subject: [3-14] When producing formatted output in Lisp, where should you put the newlines (e.g., before or after the line, FRESH-LINE vs TERPRI, ~& vs ~% in FORMAT)? Where possible, it is desirable to write functions that produce output as building blocks. In contrast with other languages, which either conservatively force a newline at various times or require the program to keep track of whether it needs to force a newline, the Lisp I/O system keeps track of whether the most recently printed character was a newline or not. The function FRESH-LINE outputs a newline only if the stream is not already at the beginning of a line. TERPRI forces a newline irrespective of the current state of the stream. These correspond to the ~& and ~% FORMAT directives, respectively. (If the Lisp I/O system can't determine whether it's physically at the beginning of a line, it assumes that a newline is needed, just in case.) Thus, if you want formatted output to be on a line of its own, start it with ~& and end it with ~%. (Some people will use a ~& also at the end, but this isn't necessary, since we know a priori that we're not at the beginning of a line. The only exception is when ~& follows a ~A, to prevent a double newline when the argument to the ~A is a formatted string with a newline at the end.) For example, the following routine prints the elements of a list, N elements per line, and then prints the total number of elements on a new line: (defun print-list (list &optional (elements-per-line 10)) (fresh-line) (loop for i upfrom 1 for element in list do (format t "~A ~:[~;~%~]" element (zerop (mod i elements-per-line)))) (format t "~&~D~%" (length list)))
Subject: [3-15] I'm using DO to do some iteration, but it doesn't terminate. Your code probably looks something like (do ((sublist list (cdr list)) ..) ((endp sublist) ..) ..) or maybe (do ((index start (+ start 2)) ..) ((= index end) ..) ..) The problem is caused by the (cdr list) and the (+ start 2) in the first line. You're using the original list and start index instead of the working sublist or index. Change them to (cdr sublist) and (+ index 2) and your code should start working.
Subject: [3-16] My program works when interpreted but not when compiled! Look for problems with your macro definitions, such as a macro that is missing a quote. When compiled, this definition essentially becomes a constant. But when interpreted, the body of the macro is executed each time the macro is called. For example, in Allegro CL the following code will behave differently when interpreted and compiled: (defvar x 10) (defmacro foo () (incf x)) (defun bar () (+ (foo) (foo))) Putting a quote before the (incf x) in the definition of foo fixes the problem. If you use (SETF (SYMBOL-FUNCTION 'foo) ...) to change the definition of a built-in Lisp function named FOO, be aware that this may not work correctly (i.e., as desired) in compiled code in all Lisps. In some Lisps, the compiler treats certain symbols in the LISP package specially, ignoring the function definition. If you want to redefine a standard function try proclaiming/declaring it NOTINLINE prior to compiling any use that should go through the function cell. (Note that this is not guarranteed to work, since X3J13 has stated that it is not permitted to redefine any of the standard functions). ---------------------------------------------------------------- ;;; *EOF*

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Last Update March 27 2014 @ 02:11 PM