According to CLHS entry for the INTERSECTION function (http://www.lispworks.com/documentation/HyperSpec/Body/f_isec_.htm):
For every pair that satifies the test, exactly one of the two elements
of the pair will be put in the result.
My problem is that i need to know which one of the two elements of the pair will be put in the result, which matters when, for example :key #'car is used to extract the arguments to test against, since the cdr might be different. I would like to have a guarantee that either always the first or always the second element will be put in the result. Am I missing something or is this just unspecified behaviour, so I shouldnt use it for my case?
intersection simply doesn't make the guarantee you want; it implements set-theoretic intersection with a lot of extras, but just not that extra. You'll have to roll your own.
Related
I'd like to compare the code contents of two syntax objects and ignore things like contexts. Is converting them to datum the only way to do so? Like:
(equal? (syntax->datum #'(x+1)) (syntax->datum #'(x+1)))
If you want to compare both objects without deconstructing them at all, then yes.
HOWEVER, the problem with this method is that it only compares the datum attached to two syntax objects, and won't actually compare their binding information.
The analogy that I've heard (from Ryan Culpepper), is this is kind of like taking two paintings, draining of them of their color, and seeing if they are identical. While they might be similar in some ways, you will miss a lot of differences from the different colors.
A better approach (although it does require some work), is to use syntax-e to destruct the syntax object into more primitive lists of syntax objects, and do this until you get identifiers (basically a syntax object whose datum is a symbol), from there, you can generally use free-identifier=? (and sometimes bound-identifier=? to see if each identifier can bind each other, and identifier-binding to compare module level identifiers.
The reason why there isn't a single simple predicate to compare two arbitrary syntax objects is because, generally, there isn't really one good definition for what makes two pieces of code equal, even if you only care about syntactic equality. For example, using the functions referenced above doesn't track internal bindings in a syntax object, so you will still get a very strict definition of what it means to be 'equal'. that is, both syntax objects have the same structure with identifiers that are either bound to the same module, or are free-identifier=?.
As such, before you use this answer, I highly recommend you take a step back and make sure this is really what you want to do. Once in a blue moon it is, but most of the time you actually are trying to solve a similar, yet simpler, problem.
Here's a concrete example of one possible way you could do the "better approach" Leif Andersen mentioned.
I have used this in multiple places for testing purposes, though if anyone wanted to use it in non-test code, they would probably want to re-visit some of the design decisions.
However, things like the equal?/recur pattern used here should be helpful no matter how you decide to define what equality means.
Some of the decisions you might want to make different choices on:
On identifiers, do you want to check that the scopes are exactly the same (bound-identifier=?), or would you want to assume that they would be bound outside of the syntax object and check that they are bound to the same thing, even if they have different scopes (free-identifier=?)? Note that if you choose the first one, then checking the results of macro expansion will sometimes return #false because of scope differences, but if you choose the second one, then if any identifier is not bound outside of the syntax object, then it would be as if you only care about symbol=? equality on names, so it will return #true in some places where it shouldn't. I chose the first one bound-identifier=? here because for testing, a "false positive" where the test fails is better than a "false negative" where the tests succeeds in cases it shouldn't.
On source locations, do you want to check that they are equal, or do you want to ignore them? This code ignores them because it's only for testing purposes, but if you want equality only for things which have the same source location, you might want to check that using functions like build-source-location-list.
On syntax properties, do you want to check that they are equal, or do you want to ignore them? This code ignores them because it's only for testing purposes, but if you want to check that you might want to use functions like syntax-property-symbol-keys.
Finally here is the code. It may not be exactly what you want depending on how you answered the questions above. However, its structure and how it uses equal?/recur might be helpful to you.
(require rackunit)
;; Works on fully wrapped, non-wrapped, and partially
;; wrapped values, and it checks that the inputs
;; are wrapped in all the same places. It checks scopes,
;; but it does not check source location.
(define-binary-check (check-stx=? stx=? actual expected))
;; Stx Stx -> Bool
(define (stx=? a b)
(cond
[(and (identifier? a) (identifier? b))
(bound-identifier=? a b)]
[(and (syntax? a) (syntax? b))
(and (bound-identifier=? (datum->syntax a '||) (datum->syntax b '||))
(stx=? (syntax-e a) (syntax-e b)))]
[else
(equal?/recur a b stx=?)]))
The following PL code does not work under #lang pl:
Edited code according to Alexis Kings answer
(define-type BINTREE
[Leaf Number]
[Node BINTREE BINTREE])
(: retrieve-leaf : BINTREE -> Number)
(define (retrieve-leaf btree)
(match btree
[(Leaf number) number])
What i'd like to achieve is as follows:
Receive a BINTREE as input
Check whether the tree is simply a leaf
Return the leaf numerical value
This might be a basic question but how would I go about solving this?
EDIT: The above seems to work if cases is used instead of match.
Why is that?
As you've discovered, match and cases are two similar but separate
things. The first is used for general Racket values, and the second is
used for things that you defined with define-type. Unfortunately,
they don't mix well in either direction, so if you have a defined type
then you need to use cases.
As for the reason for that, it's kind of complicated... One thing is
that the pl language was made well before match was powerful enough
to deal with arbitrary values conveniently. It does now, but it cannot
be easily tweaked to do what cases does: the idea behind define-type
is to make programming simple by making it mandatory to use just
cases for such values --- there are no field accessors, no predicates
for the variants (just for the whole type), and certainly no mutation.
Still, it is possible to do anything you need with just cases. If you
read around, the core idea is to mimic disjoint union types in HM
languages like ML and Haskell, and with only cases pattern matching
available, many functions are easy to start since there's a single way
to deal with them.
match and Typed Racket got closer to being able to do these things,
but it's still not really powerful enough to do all of that --- which is
why cases will stay separate from match in the near future.
As a side note, this is in contrast to what I want --- I know that this
is often a point of confusion, so I'd love to have just match used
throughout. Maybe I'll break at some point and hack things so that
cases is also called match, and the contents of the branches would
be used to guess if you really need the real match or the cases
version. But that would really be a crude hack.
I think you're on the right track, but your match syntax isn't correct. It should look like this:
(: retrieve-leaf : BINTREE -> Number)
(define (retrieve-leaf btree)
(match btree
[(Leaf number) number]))
The match pattern clauses must be inside the match form. Additionally, number is just a binding, not a procedure, so it doesn't need to be in parens.
I am defining a variable in the beginning of my source code in MATLAB. Now I would like to know at which lines this variable effects something. In other words, I would like to see all lines in which that variable is read out. This wish does not only include all accesses in the current function, but also possible accesses in sub-functions that use this variable as an input argument. In this way, I can see in a quick way where my change of this variable takes any influence.
Is there any possibility to do so in MATLAB? A graphical marking of the corresponding lines would be nice but a command line output might be even more practical.
You may always use "Find Files" to search for a certain keyword or expression. In my R2012a/Windows version is in Edit > Find Files..., with the keyboard shortcut [CTRL] + [SHIFT] + [F].
The result will be a list of lines where the searched string is found, in all the files found in the specified folder. Please check out the options in the search dialog for more details and flexibility.
Later edit: thanks to #zinjaai, I noticed that #tc88 required that this tool should track the effect of the name of the variable inside the functions/subfunctions. I think this is:
very difficult to achieve. The problem of running trough all the possible values and branching on every possible conditional expression is... well is hard. I think is halting-problem-hard.
in 90% of the case the assumption that the output of a function is influenced by the input is true. But the input and the output are part of the same statement (assigning the result of a function) so looking for where the variable is used as argument should suffice to identify what output variables are affected..
There are perverse cases where functions will alter arguments that are handle-type (because the argument is not copied, but referenced). This side-effect will break the assumption 2, and is one of the main reasons why 1. Outlining the cases when these side effects take place is again, hard, and is better to assume that all of them are modified.
Some other cases are inherently undecidable, because they don't depend on the computer states, but on the state of the "outside world". Example: suppose one calls uigetfile. The function returns a char type when the user selects a file, and a double type for the case when the user chooses not to select a file. Obviously the two cases will be treated differently. How could you know which variables are created/modified before the user deciding?
In conclusion: I think that human intuition, plus the MATLAB Debugger (for run time), and the Find Files (for quick search where a variable is used) and depfun (for quick identification of function dependence) is way cheaper. But I would like to be wrong. :-)
I'm trying to implement a Towers of Hanoi recursively with Common Lisp. I know what the recursive calls are and how they work, but I'm just lost with how I would go about moving something from the end of one list, to the end of another list. I was trying to do some research on how to do this, but I couldn't find anything online.
Any help would be greatly appreciated.
Thanks!
You can remove the last element of a list with butlast, get last element with last, and you cat append a list to another list with append (you just cons your element to be added so that it's a one element list). Working on the end of lists in CL are not optimal as every function needs to traverse the list in order to find the last one, but it is done when you need to add/remove in both ends.
With Tower of Hanoi you are stacking disks on top of each other and the last one put is the first one out. In Common Lisp (in fact any Lisp cousin) you can easily add to front and remove from front with doing (cdr pole-a) in the recursive call to remove the top element from pole-a and add to pole-b with (cons (car pole-a) pole-b) in the recursive call.
I imagine you need this to actually be able to see which disk is moved in each stage, since you need no such structure to calculate the moves needed. In such a scenario you only need the names of the poles and the level the recursion and the number of disks you want to move in this round.
Ok I need some help with thinking through this conceputally.
I need to check if a list and another list is structurally equal.
For example:
(a (bc) de)) is the same as (f (gh) ij)), because they have the same structure.
Now cleary the base case will be if both list are empty they are structurally equal.
The recursive case on the other hand I'm not sure where to start.
Some ideas:
Well we are not going to care if the elements are == to each other because that doesn't matter. We just care in the structure. I do know we will car down the list and recursively call the function with the cdr of the list.
The part that confuses me is how do you determine wheter an atom or sublist has the same structure?
Any help will be appreciated.
You're getting there. In the (free, online, excellent) textbook, this falls into section 17.3, "Processing two lists simultaneously: Case 3". I suggest you take a look.
http://www.htdp.org/2003-09-26/Book/curriculum-Z-H-1.html#node_toc_node_sec_17.3
One caveat: it looks like the data definition you're working with is "s-expression", which you can state like this:
;; an s-expression is either
;; - the empty list, or
;; - (cons symbol s-expression), or
;; - (cons s-expression s-expression)
Since this data definition has three cases, there are nine possibilities when considering two of them.
John Clements
(Yes, you could reduce the number of cases by embedding the data in the more general one that includes improper lists. Doesn't sound like a good idea to me.)