In Lisp, how do you get the value of an expression? - lisp

This will print data, but I want it to print show. I want to print the value, not the expression, how would I do it?
(defun display (x)
(list x))
(setq temp 'data)
(set temp 'show)
(display 'data)
what if u dont know if the variable is bound or not? i have to write a function that take a key and a value, if key does not exist, then i have to do setq key value, if the key already exist, then i would add the value to the key. In this case, if i do (storedata key value), if value have not been bounded, i get an unbound error, how would i handle this case?
for example, if there is no mydata and i do (storedata value mydata) then mydata would become (value), now if i do (storedata value2 mydata) then mydata becomes (value value2).

Quoting a list or a symbol in Lisp with ' is exactly equivalent to using the special form (quote ...). It's specifically for making the quoted thing not get evaluated. 'data in Lisp code or typed into the REPL is the same thing as (quote data), and evaluates to the symbol data.
data without the quote evaluates to the value of the variable data in the current scope. So, at the REPL:
[1]> (setq data 14)
14
[2]> data
14
The first expression also evaluates to 14 because setq returns the value of the bound variable (in this respect acting like the assignment operator = in C).
What you've done in the above code is to set the variable named temp to contain the symbol data, then, by using set (without the setq), set the variable named data to the symbol show. This is a bit similar to using soft references in Perl (for example), but I don't think it's particularly widely used or advisable as a Lisp technique.
By the way, your display procedure is probably not doing what you think either: it returns a single element list of whatever you pass to it. The fact that the value gets printed when you type it into the REPL is just because the value of any expression gets printed at the REPL. To display a value in a program you might use print or maybe format. (I'm assuming you're using Common Lisp, since it's obviously not Scheme, but maybe it's some other Lisp variety, in which case that link won't help.)

You are quoting data. If you want it to be evaluated you should just call
(display data)

Related

What do common lisp function/special form/macro/etc. names mean and where can I find this information?

When I was learning HTML it was very helpful for me to know that ol means ordered list, tr is table row, etc. Some of the lisp primitives/forms are easy: funcall should be function call, defmacro - define macro. Some are in the middle - incf is... increment... f??? But because common lisp is so old, this primitives/special forms/etc... don't seem to ring a bell. Can you guys, help me with figuring them out? And even more importantly: Where can I find an authoritative resource on learning the meaning/history behind each and every one of them? (I will accept an answer based on this second question)
The documentation doesn't help me too:
* (describe #'let)
#<CLOSURE (:SPECIAL LET) {10013DC6AB}>
[compiled closure]
Lambda-list: (&REST ARGS)
Derived type: (FUNCTION (&REST T) NIL)
Documentation:
T
Source file: SYS:SRC;COMPILER;INFO-FUNCTIONS.LISP
* (documentation 'let 'function)
"LET ({(var [value]) | var}*) declaration* form*
During evaluation of the FORMS, bind the VARS to the result of evaluating the
VALUE forms. The variables are bound in parallel after all of the VALUES forms
have been evaluated."
* (inspect 'let)
The object is a SYMBOL.
0. Name: "LET"
1. Package: #<PACKAGE "COMMON-LISP">
2. Value: "unbound"
3. Function: #<CLOSURE (:SPECIAL LET) {10013DC6AB}>
4. Plist: (SB-WALKER::WALKER-TEMPLATE SB-WALKER::WALK-LET)
What do the following lisp primitives/special forms/special operators/functions mean?
let, flet
progn
car
cdr
acc
setq, setf
incf
(write more in the comments so we can make a good list!)
let: LET variable-name be bound to a certain value
flet: LET Function-name be bound to a certain function
progn: execute a PROGram sequence and return the Nth value (the last value)
car: Contents of the Address Register (historic)
cdr: Contents of the Decrement Register (historic)
acc: ACCumulator
setq: SET Quote, a variant of the set function, where in setq the user doesn't quote the variable
setf: SET Function quoted, shorter name of the original name setfq. Here the function/place is not evaluated.
incf: INCrement Function quoted, similar to setf. Increments a place.
Other conventions:
Macros / Special Forms who change a place should have an f at the end: setf, psetf, incf, decf, ...
Macros who are DEFining something should have def or define in front: defun, defmethod, defclass, define-method-combination...
Functions who are destructive should have an n in front, for Non-consing: nreverse, ...
Predicates have a p or -p at the end: adjustable-array-p, alpha-char-p,...
special variables have * at the front and back: *standard-output*, ...
There are other naming conventions.
let: Well, that's a normal word, and is used like in maths ("let x = 3 in ...").
flet: I'd guess "function let", because that's what it does.
progn: This is probably related also to prog1 and prog2. I read is as "program whose nth form dictates the result value". The program has n forms, so it's the last one that forms the result value of the progn form.
car and cdr: "Contents address register" resp. "Contents decrement register". This is related to the IBM 704 which Lisp was originally implemented for.
setq: "set quote", originally an abbreviation for (set (quote *abc*) value).
setf: "set field", came up when lexical variables appeared. This is a good read on the set functions.
Where can I find an authoritative resource on learning the meaning/history behind each and every one of them?
The HyperSpec is a good place to start, and ultimately the ANSI standard. Though "Common Lisp The Language" could also shine some light on the history of some names.
(Oh, and you got defmacro wrong, that's "Definition for Mac, Read Only." ;) )

Bizarre quoted list example from On Lisp

This passage from On Lisp is genuinely confusing -- it is not clear how returning a quoted list such as '(oh my) can actually alter how the function behaves in the future: won't the returned list be generated again in the function from scratch, the next time it is called?
If we define exclaim so that its return value
incorporates a quoted list,
(defun exclaim (expression)
(append expression ’(oh my)))
Then any later destructive modification of the return value
(exclaim ’(lions and tigers and bears))
-> (LIONS AND TIGERS AND BEARS OH MY)
(nconc * ’(goodness))
-> (LIONS AND TIGERS AND BEARS OH MY GOODNESS)
could alter the list within the function:
(exclaim ’(fixnums and bignums and floats))
-> (FIXNUMS AND BIGNUMS AND FLOATS OH MY GOODNESS)
To make exclaim proof against such problems, it should be written:
(defun exclaim (expression)
(append expression (list ’oh ’my)))
How exactly is that last call to exclaim adding the word goodness to the result? The function is not referencing any outside variable so how did the separate call to nconc actually alter how the exclaim function works?
a) the effects of modifying literal lists is undefined in the Common Lisp standard. What you here see as an example is one possible behavior.
(1 2 3 4) is a literal list. But a call to LIST like in (list 1 2 3 4) returns a freshly consed list at runtime.
b) the list is literal data in the code of the function. Every call will return exactly this data object. If you want to provide a fresh list on each call, then you need to use something like LIST or COPY-LIST.
c) Since the returned list is always the same literal data object, modifying it CAN have this effect as described. One could imagine also that an error happens if the code and its objects are allocated in a read-only memory. Modifying the list then would try to write to read-only memory.
d) One thing to keep in mind when working with literal list data in source code is this: the Lisp compiler is free to optimize the storage. If a list happens to be multiple times in the source code, a compiler is allowed to detect this and to only create ONE list. All the various places would then point to this one list. Thus modifying the list would have the effect, that these changes could be visible in several places.
This may also happen with other literal data objects like arrays/vectors.
If your data structure is a part of the code, you return this internal data structure, you modify this data structure - then you try to modify your code.
Note also that Lisp can be executed by an Interpreter. The interpreter typically works on the Lisp source structure - the code is not machine code, but interpreted Lisp code as Lisp data. Here you might be able to modify the source code at runtime, not only the data embedded in the source code.

Evaluating arguments of Clojure tagged literals

I have been trying to use Clojure tagged literals, and noticed that the reader does not evaluate the arguments, very much like a macro. This makes sense, but what is the appropriate solution for doing this? Explicit eval?
Example: given this function
(defn my-data
([[arg]]
(prn (symbol? arg))
:ok))
and this definition data_readers.clj
{myns/my-data my.ns/my-data}
The following behave differently:
> (let [x 1] (my.ns/my-data [x]))
false
:ok
So the x passed in is evaluated before being passed into my-data. On the other hand:
> (let [x 1] #myns/my-data [x])
true
:ok
So if I want to use the value of x inside my-data, the my-data function needs to do something about it, namely check that x is a symbol, and if so, use (eval x). That seems ugly. Is there a better approach?
Summary
There is no way to get at the value of the local x in your example, primarily because locals only get assigned values at runtime, whereas tagged literals are handled at read time. (There's also compile time in between; it is impossible to get at locals' values at compile time; therefore macros cannot get at locals' values either.)1
The better approach is to use a regular function at runtime, since after all you want to construct a value based on the runtime values of some parameters. Tagged literals really are literals and should be used as such.
Extended discussion
To illustrate the issue described above:
(binding [*data-readers* {'bar (fn [_] (java.util.Date.))}]
(eval (read-string "(defn foo [] #bar x)")))
foo will always return the same value, because the reader has only one opportunity to return a value for #bar x which is then baked into foo's bytecode.
Notice also that instead of passing it to eval directly, we could store the data structure returned by the call to read-string and compile it at an arbitrary point in the future; the value returned by foo would remain the same. Clearly there's no way for such a literal value to depend on the future values of any locals; in fact, during the reader's operation, it is not even clear which symbols will come to name locals -- that's for the compiler to determine and the result of that determination may be non-obvious in cases where macros are involved.
Of course the reader is free to return a form which looks like a function call, the name of a local etc. To show one example:
(binding [*data-readers* {'bar (fn [sym] (list sym 1 2 3 4 5))}]
(eval (read-string "#bar *")))
;= 120
;; substituting + for * in the string yields a value of 15
Here #bar f becomes equivalent to (f 1 2 3 4 5). Needless to say, this is an abuse of notation and doesn't really do what you asked for.
1 It's worth pointing out that eval has no access to locals (it always operates in the global scope), but the issue with locals not having values assigned before runtime is more fundamental.

What's the return value of `define` in Scheme?

I'm curious about the return value of define in Scheme. So I wrote the following lines in Racket
#lang r5rs
(display (define a 3))
And get the error
define: not allowed in an expression context in: (define a 3)
I have 2 questions about this:
Does it mean that define has no return value?
According to R5RS, define is not an expression. It's a program structure. Is it true that only expressions have return values, and other forms don't?
"If a tree falls in a forest and no one is around to hear it, does it make a sound?"
It's not valid to use define in any context where a return value could meaningfully be obtained. So it's moot whether it has a return value or not; you'll never be able to observe it.
In Scheme, define can only be used in two places:
At the top level, or
At the very beginning of a "body".
In neither of those places is a "return value" relevant.

Strange Lisp Quoting scenario - Graham's On Lisp, page 37

I'm working my way through Graham's book "On Lisp" and can't understand the following example at page 37:
If we define exclaim so that its return value
incorporates a quoted list,
(defun exclaim (expression)
(append expression ’(oh my)))
> (exclaim ’(lions and tigers and bears))
(LIONS AND TIGERS AND BEARS OH MY)
> (nconc * ’(goodness))
(LIONS AND TIGERS AND BEARS OH MY GOODNESS)
could alter the list within the function:
> (exclaim ’(fixnums and bignums and floats))
(FIXNUMS AND BIGNUMS AND FLOATS OH MY GOODNESS)
To make exclaim proof against such problems, it should be written:
(defun exclaim (expression)
(append expression (list ’oh ’my)))
Does anyone understand what's going on here? This is seriously screwing with my mental model of what quoting does.
nconc is a destructive operation that alters its first argument by changing its tail. In this case, it means that the constant list '(oh my) gets a new tail.
To hopefully make this clearer. It's a bit like this:
; Hidden variable inside exclaim
oh_my = oh → my → nil
(exclaim '(lions and tigers and bears)) =
lions → and → tigers → and → bears → oh_my
(nconc * '(goodness)) destructively appends goodness to the last result:
lions → and → tigers → and → bears → oh → my → goodness → nil
so now, oh_my = oh → my → goodness → nil
Replacing '(oh my) with (list 'oh 'my) fixes this because there is no longer a constant being shared by all and sundry. Each call to exclaim generates a new list (the list function's purpose in life is to create brand new lists).
The observation that your mental model of quoting may be flawed is an excellent one—although it may or may not apply depending on what that mental model is.
First, remember that there are various stages to program execution. A Lisp environment must first read the program text into data structures (lists, symbols, and various literal data such as strings and numbers). Next, it may or may not compile those data structures into machine code or some sort of intermediary format. Finally, the resulting code is evaluated (in the case of machine code, of course, this may simply mean jumping to the appropriate address).
Let's put the issue of compilation aside for now and focus on the reading and evaluation stages, assuming (for simplicity) that the evaluator's input is the list of data structures read by the reader.
Consider a form (QUOTE x) where x is some textual representation of an object. This may be symbol literal as in (QUOTE ABC), a list literal as in (QUOTE (A B C)), a string literal as in (QUOTE "abc"), or any other kind of literal. In the reading stage, the reader will read the form as a list (call it form1) whose first element is the symbol QUOTE and whose second element is the object x' whose textual representation is x. Note that I'm specifically saying that the object x' is stored within the list that represents the expression, i.e. in a sense, it's stored as a part of the code itself.
Now it's the evaluator's turn. The evaluator's input is form1, which is a list. So it looks at the first element of form1, and, having determined that it is the symbol QUOTE, it returns as the result of the evaluation the second element of the list. This is the crucial point. The evaluator returns the second element of the list to be evaluated, which is what the reader read in in the first execution stage (prior to compilation!). That's all it does. There's no magic to it, it's very simple, and significantly, no new objects are created and no existing ones are copied.
Therefore, whenever you modify a “quoted list”, you're modifying the code itself. Self-modifying code is a very confusing thing, and in this case, the behaviour is actually undefined (because ANSI Common Lisp permits implementations to put code in read-only memory).
Of course, the above is merely a mental model. Implementations are free to implement the model in various ways, and in fact, I know of no implementation of Common Lisp that, like my explanation, does no compilation at all. Still, this is the basic idea.
In Common Lisp.
Remember:
'(1 2 3 4)
Above is a literal list. Constant data.
(list 1 2 3 4)
LIST is a function that when call returns a fresh new list with its arguments as list elements.
Avoid modifying literal lists. The effects are not standardized. Imagine a Lisp that compiles all constant data into a read only memory area. Imagine a Lisp that takes constant lists and shares them across functions.
(defun a () '(1 2 3)
(defun b () '(1 2 3))
A Lisp compiler may create one list that is shared by both functions.
If you modify the list returned by function a
it might not be changed
it might be changed
it might be an error
it might also change the list returned by function b
Implementations have the freedom to do what they like. This leaves room for optimizations.