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  • ManyToManyField error when having recursive structure. How to solve it?

    - by luc
    Hello, I have the following table in the model with a recursive structure (a page can have children pages) class DynamicPage(models.Model): name = models.CharField("Titre",max_length=200) parent = models.ForeignKey('self',null=True,blank=True) I want to create another table with manytomany relation with this one: class UserMessage(models.Model): name = models.CharField("Nom", max_length=100) page = models.ManyToManyField(DynamicPage) The generated SQL creates the following constraint: ALTER TABLE `website_dynamicpage` ADD CONSTRAINT `parent_id_refs_id_29c58e1b` FOREIGN KEY (`parent_id`) REFERENCES `website_dynamicpage` (`id`); I would like to have the ManyToMany with the page itself (the id) and not with the parent field. How to modify the model to make the constraint using the id and not the parent? Thanks in advance

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  • Recursive MySQL function call eats up too much memory and dies.

    - by kylex
    I have the following recursive function which works... up until a point. Then the script asks for more memory once the queries exceed about 100, and when I add more memory, the script typically just dies (I end up with a white screen on my browser). public function returnPArray($parent=0,$depth=0,$orderBy = 'showOrder ASC'){ $query = mysql_query("SELECT *, UNIX_TIMESTAMP(lastDate) AS whenTime FROM these_pages WHERE parent = '".$parent."' AND deleted = 'N' ORDER BY ".$orderBy.""); $rows = mysql_num_rows($query); while($row = mysql_fetch_assoc($query)){ // This uses my class and places the content in an array. MyClass::$_navArray[] = array( 'id' => $row['id'], 'parent' => $row['parent'] ); MyClass::returnPArray($row['id'],($depth+1)); } $i++; } Can anyone help me make this query less resource intensive?

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  • jQuery: is there a way to make a recursive child selector?

    - by gsquare567
    instead of checking only the immediate children of an element, i would like to recursively check all children of the element. specifically, something like $("#survey11Form>input[type=text]:visible").val(); with the html: <form id="survey11Form" name="survey11Form" action="#" method="post"> <div id="survey11Div"> <fieldset> <legend>TEST'TEST"TEST</legend> <div class="label"> <label test="" title="TEST'TEST" for="answer15"> TEST'TEST"TEST </label></div> <div class="fieldWrapper text required"> <div style="width: 146px;" class="cellValue"> <input type="text" title="TEST'TEST" value="" id="survey11answer15" name="survey11answer15"> should give me the value of that input. the jquery i've come up with does not. any ideas as to what would work in this situation (and all recursive situations)? thanks!

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  • I'm having trouble with using std::stack to retrieve the values from a recursive function.

    - by Peter Stewart
    Thanks to the help I received in this post: http://stackoverflow.com/questions/2761918/how-do-i-use-this-in-a-member-function I have a nice, concise recursive function to traverse a tree in postfix order: void Node::postfix() { if (left != __nullptr) { left->postfix(); } if (right != __nullptr) { right->postfix(); } cout<<cargo<<"\n"; return; }; Now I need to evaluate the values and operators as they are returned. My problem is how to retrieve them. I tried the std::stack: #include <stack> stack <char*> s; void Node::postfix() { if (left != __nullptr) { left->postfix(); } if (right != __nullptr) { right->postfix(); } s.push(cargo); return; }; but when I tried to access it in main() while (!s.empty()) { cout<<s.top<<"\n"; s.pop; } I got the error: 'std::stack<_Ty::top': function call missing argument list; use '&std::stack<_Ty::top' to create a pointer to member' I'm stuck. Another question to follow shortly.

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  • Can Haskell's Parsec library be used to implement a recursive descent parser with backup?

    - by Thor Thurn
    I've been considering using Haskell's Parsec parsing library to parse a subset of Java as a recursive descent parser as an alternative to more traditional parser-generator solutions like Happy. Parsec seems very easy to use, and parse speed is definitely not a factor for me. I'm wondering, though, if it's possible to implement "backup" with Parsec, a technique which finds the correct production to use by trying each one in turn. For a simple example, consider the very start of the JLS Java grammar: Literal: IntegerLiteral FloatingPointLiteral I'd like a way to not have to figure out how I should order these two rules to get the parse to succeed. As it stands, a naive implementation like this: literal = do { x <- try (do { v <- integer; return (IntLiteral v)}) <|> (do { v <- float; return (FPLiteral v)}); return(Literal x) } Will not work... inputs like "15.2" will cause the integer parser to succeed first, and then the whole thing will choke on the "." symbol. In this case, of course, it's obvious that you can solve the problem by re-ordering the two productions. In the general case, though, finding things like this is going to be a nightmare, and it's very likely that I'll miss some cases. Ideally, I'd like a way to have Parsec figure out stuff like this for me. Is this possible, or am I simply trying to do too much with the library? The Parsec documentation claims that it can "parse context-sensitive, infinite look-ahead grammars", so it seems like something like I should be able to do something here.

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  • Should not a tail-recursive function also be faster?

    - by Balint Erdi
    I have the following Clojure code to calculate a number with a certain "factorable" property. (what exactly the code does is secondary). (defn factor-9 ([] (let [digits (take 9 (iterate #(inc %) 1)) nums (map (fn [x] ,(Integer. (apply str x))) (permutations digits))] (some (fn [x] (and (factor-9 x) x)) nums))) ([n] (or (= 1 (count (str n))) (and (divisible-by-length n) (factor-9 (quot n 10)))))) Now, I'm into TCO and realize that Clojure can only provide tail-recursion if explicitly told so using the recur keyword. So I've rewritten the code to do that (replacing factor-9 with recur being the only difference): (defn factor-9 ([] (let [digits (take 9 (iterate #(inc %) 1)) nums (map (fn [x] ,(Integer. (apply str x))) (permutations digits))] (some (fn [x] (and (factor-9 x) x)) nums))) ([n] (or (= 1 (count (str n))) (and (divisible-by-length n) (recur (quot n 10)))))) To my knowledge, TCO has a double benefit. The first one is that it does not use the stack as heavily as a non tail-recursive call and thus does not blow it on larger recursions. The second, I think is that consequently it's faster since it can be converted to a loop. Now, I've made a very rough benchmark and have not seen any difference between the two implementations although. Am I wrong in my second assumption or does this have something to do with running on the JVM (which does not have automatic TCO) and recur using a trick to achieve it? Thank you.

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  • Recursive breadth-first travel function in Java or C++?

    - by joejax
    Here is a java code for breadth-first travel: void breadthFirstNonRecursive(){ Queue<Node> queue = new java.util.LinkedList<Node>(); queue.offer(root); while(!queue.isEmpty()){ Node node = queue.poll(); visit(node); if (node.left != null) queue.offer(node.left); if (node.right != null) queue.offer(node.right); } } Is it possible to write a recursive function to do the same? At first, I thought this would be easy, so I came out with this: void breadthFirstRecursive(){ Queue<Node> q = new LinkedList<Node>(); breadthFirst(root, q); } void breadthFirst(Node node, Queue<Node> q){ if (node == null) return; q.offer(node); Node n = q.poll(); visit(n); if (n.left != null) breadthFirst(n.left, q); if (n.right != null) breadthFirst(n.right, q); } Then I found it doesn't work. It is actually does the same thing as this: void preOrder(Node node) { if (node == null) return; visit(node); preOrder(node.left); preOrder(node.right); } Has any one thought about this before?

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  • How to change a recursive function for count files and catalogues?

    - by user661999
    <?php function scan_dir($dirname) { $file_count = 0 ; $dir_count = 0 ; $dir = opendir($dirname); while (($file = readdir($dir)) !== false) { if($file != "." && $file != "..") { if(is_file($dirname."/".$file)) ++$file_count; if(is_dir($dirname."/".$file)) { ++ $dir_count; scan_dir($dirname."/".$file); } } } closedir($dir); echo "There are $dir_count catalogues and $file_count files.<br>"; } $dirname = "/home/user/path"; scan_dir($dirname); ?> Hello, I have a recursive function for count files and catalogues. It returns result for each catalogue. But I need a common result. How to change the script? It returns : There are 0 catalogues and 3 files. There are 0 catalogues and 1 files. There are 2 catalogues and 14 files. I want: There are 2 catalogues and 18 files.

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  • Apply rewrite rule for all but all the files (recursive) in a subdirectory?

    - by user784637
    I have an .htaccess file in the root of the website that looks like this RewriteRule ^some-blog-post-title/ http://website/read/flowers/a-new-title-for-this-post/ [R=301,L] RewriteRule ^some-blog-post-title2/ http://website/read/flowers/a-new-title-for-this-post2/ [R=301,L] <IfModule mod_rewrite.c> RewriteEngine On ## Redirects for all pages except for files in wp-content to website/read RewriteCond %{REQUEST_FILENAME} !-f RewriteCond %{REQUEST_FILENAME} !-d RewriteCond %{REQUEST_URI} !/wp-content RewriteRule ^(.*)$ http://website/read/$1 [L,QSA] #RewriteRule ^http://website/read [R=301,L] RewriteBase / RewriteRule ^index\.php$ - [L] RewriteCond %{REQUEST_FILENAME} !-f RewriteCond %{REQUEST_FILENAME} !-d RewriteRule . /index.php [L] </IfModule> My intent is to redirect people to the new blog post location if they propose one of those special blog posts. If that's not the case then they should be redirected to http://website.com/read. Nothing from http://website.com/wp-content/* should be redirected. So far conditions 1 and 3 are being met. How can I meet condition 2?

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  • Recursive N-way merge/diff algorithm for directory trees?

    - by BobMcGee
    What algorithms or Java libraries are available to do N-way, recursive diff/merge of directories? I need to be able to generate a list of folder trees that have many identical files, and have subdirectories with many similar files. I want to be able to use 2-way merge operations to quickly remove as much redundancy as possible. Goals: Find pairs of directories that have many similar files between them. Generate short list of directory pairs that can be synchronized with 2-way merge to eliminate duplicates Should operate recursively (there may be nested duplicates of higher-level directories) Run time and storage should be O(n log n) in numbers of directories and files Should be able to use an embedded DB or page to disk for processing more files than fit in memory (100,000+). Optional: generate an ancestry and change-set between folders Optional: sort the merge operations by how many duplicates they can elliminate I know how to use hashes to find duplicate files in roughly O(n) space, but I'm at a loss for how to go from this to finding partially overlapping sets between folders and their children. EDIT: some clarification The tricky part is the difference between "exact same" contents (otherwise hashing file hashes would work) and "similar" (which will not). Basically, I want to feed this algorithm at a set of directories and have it return a set of 2-way merge operations I can perform in order to reduce duplicates as much as possible with as few conflicts possible. It's effectively constructing an ancestry tree showing which folders are derived from each other. The end goal is to let me incorporate a bunch of different folders into one common tree. For example, I may have a folder holding programming projects, and then copy some of its contents to another computer to work on it. Then I might back up and intermediate version to flash drive. Except I may have 8 or 10 different versions, with slightly different organizational structures or folder names. I need to be able to merge them one step at a time, so I can chose how to incorporate changes at each step of the way. This is actually more or less what I intend to do with my utility (bring together a bunch of scattered backups from different points in time). I figure if I can do it right I may as well release it as a small open source util. I think the same tricks might be useful for comparing XML trees though.

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  • Non recursive way to position a genogram in 2D points for x axis. Descendant are below

    - by Nassign
    I currently was tasked to make a genogram for a family consisting of siblings, parents with aunts and uncles with grandparents and greatgrandparents for only blood relatives. My current algorithm is using recursion. but I am wondering how to do it in non recursive way to make it more efficient. it is programmed in c# using graphics to draw on a bitmap. Current algorithm for calculating x position, the y position is by getting the generation number. public void StartCalculatePosition() { // Search the start node (The only node with targetFlg set to true) Person start = null; foreach (Person p in PersonDic.Values) { if (start == null) start = p; if (p.Targetflg) { start = p; break; } } CalcPositionRecurse(start); // Normalize the position (shift all values to positive value) // Get the minimum value (must be negative) // Then offset the position of all marriage and person with that to make it start from zero float minPosition = float.MaxValue; foreach (Person p in PersonDic.Values) { if (minPosition > p.Position) { minPosition = p.Position; } } if (minPosition < 0) { foreach (Person p in PersonDic.Values) { p.Position -= minPosition; } foreach (Marriage m in MarriageList) { m.ParentsPosition -= minPosition; m.ChildrenPosition -= minPosition; } } } /// <summary> /// Calculate position of genogram using recursion /// </summary> /// <param name="psn"></param> private void CalcPositionRecurse(Person psn) { // End the recursion if (psn.BirthMarriage == null || psn.BirthMarriage.Parents.Count == 0) { psn.Position = 0.0f; if (psn.BirthMarriage != null) { psn.BirthMarriage.ParentsPosition = 0.0f; psn.BirthMarriage.ChildrenPosition = 0.0f; } CalculateSiblingPosition(psn); return; } // Left recurse if (psn.Father != null) { CalcPositionRecurse(psn.Father); } // Right recurse if (psn.Mother != null) { CalcPositionRecurse(psn.Mother); } // Merge Position if (psn.Father != null && psn.Mother != null) { AdjustConflict(psn.Father, psn.Mother); // Position person in center of parent psn.Position = (psn.Father.Position + psn.Mother.Position) / 2; psn.BirthMarriage.ParentsPosition = psn.Position; psn.BirthMarriage.ChildrenPosition = psn.Position; } else { // Single mom or single dad if (psn.Father != null) { psn.Position = psn.Father.Position; psn.BirthMarriage.ParentsPosition = psn.Position; psn.BirthMarriage.ChildrenPosition = psn.Position; } else if (psn.Mother != null) { psn.Position = psn.Mother.Position; psn.BirthMarriage.ParentsPosition = psn.Position; psn.BirthMarriage.ChildrenPosition = psn.Position; } else { // Should not happen, checking in start of function } } // Arrange the siblings base on my position (left younger, right older) CalculateSiblingPosition(psn); } private float GetRightBoundaryAncestor(Person psn) { float rPos = psn.Position; // Get the rightmost position among siblings foreach (Person sibling in psn.Siblings) { if (sibling.Position > rPos) { rPos = sibling.Position; } } if (psn.Father != null) { float rFatherPos = GetRightBoundaryAncestor(psn.Father); if (rFatherPos > rPos) { rPos = rFatherPos; } } if (psn.Mother != null) { float rMotherPos = GetRightBoundaryAncestor(psn.Mother); if (rMotherPos > rPos) { rPos = rMotherPos; } } return rPos; } private float GetLeftBoundaryAncestor(Person psn) { float rPos = psn.Position; // Get the rightmost position among siblings foreach (Person sibling in psn.Siblings) { if (sibling.Position < rPos) { rPos = sibling.Position; } } if (psn.Father != null) { float rFatherPos = GetLeftBoundaryAncestor(psn.Father); if (rFatherPos < rPos) { rPos = rFatherPos; } } if (psn.Mother != null) { float rMotherPos = GetLeftBoundaryAncestor(psn.Mother); if (rMotherPos < rPos) { rPos = rMotherPos; } } return rPos; } /// <summary> /// Check if two parent group has conflict and compensate on the conflict /// </summary> /// <param name="leftGroup"></param> /// <param name="rightGroup"></param> public void AdjustConflict(Person leftGroup, Person rightGroup) { float leftMax = GetRightBoundaryAncestor(leftGroup); leftMax += 0.5f; float rightMin = GetLeftBoundaryAncestor(rightGroup); rightMin -= 0.5f; float diff = leftMax - rightMin; if (diff > 0.0f) { float moveHalf = Math.Abs(diff) / 2; RecurseMoveAncestor(leftGroup, 0 - moveHalf); RecurseMoveAncestor(rightGroup, moveHalf); } } /// <summary> /// Recursively move a person and all his/her ancestor /// </summary> /// <param name="psn"></param> /// <param name="moveUnit"></param> public void RecurseMoveAncestor(Person psn, float moveUnit) { psn.Position += moveUnit; foreach (Person siblings in psn.Siblings) { if (siblings.Id != psn.Id) { siblings.Position += moveUnit; } } if (psn.BirthMarriage != null) { psn.BirthMarriage.ChildrenPosition += moveUnit; psn.BirthMarriage.ParentsPosition += moveUnit; } if (psn.Father != null) { RecurseMoveAncestor(psn.Father, moveUnit); } if (psn.Mother != null) { RecurseMoveAncestor(psn.Mother, moveUnit); } } /// <summary> /// Calculate the position of the siblings /// </summary> /// <param name="psn"></param> /// <param name="anchor"></param> public void CalculateSiblingPosition(Person psn) { if (psn.Siblings.Count == 0) { return; } List<Person> sibling = psn.Siblings; int argidx; for (argidx = 0; argidx < sibling.Count; argidx++) { if (sibling[argidx].Id == psn.Id) { break; } } // Compute position for each brother that is younger that person int idx; for (idx = argidx - 1; idx >= 0; idx--) { sibling[idx].Position = sibling[idx + 1].Position - 1; } for (idx = argidx + 1; idx < sibling.Count; idx++) { sibling[idx].Position = sibling[idx - 1].Position + 1; } }

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  • A Begginner's Guide to SEO - The Basics

    n the world of Internet marketing, one important factor that will either make or break your bank is SEO. Just a short introduction about this famous acronym SEO, it is actually an acronym for search engine optimization.

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  • how can I code a recursive query in an Entity Framework model?

    - by Greg
    Hi, I have a model which includes NODES, and RELATIONSHIPS (that tie the nodes together, via a parent_node, child_node arrangement). Q1 - Is there any way in EF / Linq-to-entities to perform a query on nodes (e.g. context.Nodes..) to find say "all parents" or "or children" in the graph? Q2 - If there's not in Linq-to-entities, is there any other way to do this other than writing a method that manually goes through and doing it? Q3 - If manual is the only way to do it, should I be concerned about the number of database hits that will be going out to the database as the method keeps recursing through the data? Or more specifically, is there any EF caching type feature that might assist here in ensuring the method is performance from a "number of database hits" point of view? thanks thanks

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  • What is the best way to translate this recursive python method into Java?

    - by Simucal
    In another question I was provided with a great answer involving generating certain sets for the Chinese Postman Problem. The answer provided was: def get_pairs(s): if not s: yield [] else: i = min(s) for j in s - set([i]): for r in get_pairs(s - set([i, j])): yield [(i, j)] + r for x in get_pairs(set([1,2,3,4,5,6])): print x This will output the desire result of: [(1, 2), (3, 4), (5, 6)] [(1, 2), (3, 5), (4, 6)] [(1, 2), (3, 6), (4, 5)] [(1, 3), (2, 4), (5, 6)] [(1, 3), (2, 5), (4, 6)] [(1, 3), (2, 6), (4, 5)] [(1, 4), (2, 3), (5, 6)] [(1, 4), (2, 5), (3, 6)] [(1, 4), (2, 6), (3, 5)] [(1, 5), (2, 3), (4, 6)] [(1, 5), (2, 4), (3, 6)] [(1, 5), (2, 6), (3, 4)] [(1, 6), (2, 3), (4, 5)] [(1, 6), (2, 4), (3, 5)] [(1, 6), (2, 5), (3, 4)] This really shows off the expressiveness of Python because this is almost exactly how I would write the pseudo-code for the algorithm. I especially like the usage of yield and and the way that sets are treated as first class citizens. However, there in lies my problem. What would be the best way to: 1.Duplicate the functionality of the yield return construct in Java? Would it instead be best to maintain a list and append my partial results to this list? How would you handle the yield keyword. 2.Handle the dealing with the sets? I know that I could probably use one of the Java collections which implements that implements the Set interface and then using things like removeAll() to give me a set difference. Is this what you would do in that case? Ultimately, I'm looking to reduce this method into as concise and straightforward way as possible in Java. I'm thinking the return type of the java version of this method will likely return a list of int arrays or something similar. How would you handle the situations above when converting this method into Java?

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  • Does a recursive Ant task exist to recover properties from external file?

    - by Julia2020
    Hi, I ve got a problem in getting properties with ant from a properties file. With a simple target like this in my build.xml, i'd like to get at least two properties path1 and path2. I'd like to have a generic target to get this two properties.... in order to avoid modifying the build.xml (just adding a new prop) Any suggestions? Thanks in advance ! build.xml : <target name="TEST" description="test ant"> <property file="dependencies.properties"/> <svn> <export srcUrl="${path.prop}" destPath="${workspace}/rep/" /> </svn> </target> dependencies.properties : path1.prop = /path/to/src1 path2.prop = /path/to/src2

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  • Recursive multiline sed - remove beginning of file until pattern match.

    - by yaya3
    I have nested subdirectories containing html files. For each of these html files I want to delete from the top of the file until the pattern <div id="left- This is my attempt from osx's terminal: find . -name "*.html" -exec sed "s/.*?<div id=\"left-col/<div id=\"left-col/g" '{}' \; I get a lot of html output in the termainal, but no files contain the substitution or are written Thanks

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  • What is wrong with this recursive Windows CMD script? It won't do Ackermann properly

    - by boost
    I've got this code that I'm trying to get to calculate the Ackermann function so that I can post it up on RosettaCode. It almost works. I thought maybe there'd be a few batch file wizards on StackOverflow. ::echo off set depth=0 :ack if %1==0 goto m0 if %2==0 goto n0 :else set /a n=%2-1 set /a depth+=1 call :ack %1 %n% set t=%errorlevel% set /a depth-=1 set /a m=%1-1 set /a depth+=1 call :ack %m% %t% set t=%errorlevel% set /a depth-=1 if %depth%==0 ( exit %t% ) else ( exit /b %t% ) :m0 set/a n=%2+1 if %depth%==0 ( exit %n% ) else ( exit /b %n% ) :n0 set /a m=%1-1 set /a depth+=1 call :ack %m% %2 set t=%errorlevel% set /a depth-=1 if %depth%==0 ( exit %t% ) else ( exit /b %t% ) I use this script to test it @echo off cmd/c ackermann.cmd %1 %2 echo Ackermann of %1 %2 is %errorlevel% A sample output, for Test 1 1, gives: >test 1 1 >set depth=0 >if 1 == 0 goto m0 >if 1 == 0 goto n0 >set /a n=1-1 >set /a depth+=1 >call :ack 1 0 >if 1 == 0 goto m0 >if 0 == 0 goto n0 >set /a m=1-1 >set /a depth+=1 >call :ack 0 0 >if 0 == 0 goto m0 >set/a n=0+1 >if 2 == 0 (exit 1 ) else (exit /b 1 ) >set t=1 >set /a depth-=1 >if 1 == 0 (exit 1 ) else (exit /b 1 ) >set t=1 >set /a depth-=1 >set /a m=1-1 >set /a depth+=1 >call :ack 0 1 >if 0 == 0 goto m0 >set/a n=1+1 >if 1 == 0 (exit 2 ) else (exit /b 2 ) >set t=2 >set /a depth-=1 >if 0 == 0 (exit 2 ) else (exit /b 2 ) Ackermann of 1 1 is 2

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  • How can I perform an idiomatic non-recursive flatten in ruby?

    - by nasmorn
    I have a method that returns an array of arrays. For convenience I use collect on a collection to gather them together. arr = collection.collect {|item| item.get_array_of_arrays} Now I would like to have a single array that contains all the arrays. Of course I can loop over the array and use the + operator to do that. newarr = [] arr.each {|item| newarr += item} But this is kind of ugly, is there a better way?

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  • Where should I initialize variables for an OO Recursive Descent Parse Tree?

    - by Vasto
    I'd like to preface this by stating that this is for a class, so please don't solve this for me. One of my labs for my cse class is creating an interpreter for a BNF that was provided. I understand most of the concepts, but I'm trying to build up my tree and I'm unsure where to initialize values. I've tried in both the constructor, and in the methods but Eclipse's debugger still only shows the left branch, even though it runs through completely. Here is my main procedure so you can get an idea of how I'm calling the methods. public class Parser { public static void main(String[] args) throws IOException { FileTokenizer instance = FileTokenizer.Instance(); FileTokenizer.main(args); Prog prog = new Prog(); prog.ParseProg(); prog.PrintProg(); prog.ExecProg(); } Now here is My Prog class: public class Prog { private DeclSeq ds; private StmtSeq ss; Prog() { ds = new DeclSeq(); ss = new StmtSeq(); } public void ParseProg() { FileTokenizer instance = FileTokenizer.Instance(); instance.skipToken(); //Skips program (1) // ds = new DeclSeq(); ds.ParseDS(); instance.skipToken(); //Skips begin (2) // ss = new StmtSeq(); ss.ParseSS(); instance.skipToken(); } I've tried having Prog() { ds = null; ss = null; } public void ParseProg() { FileTokenizer instance = FileTokenizer.Instance(); instance.skipToken(); //Skips program (1) ds = new DeclSeq(); ds.ParseDS(); ... But it gave me the same error. I need the parse tree built up so I can do a pretty print and an execute command, but like I said, I only get the left branch. Any help would be appreciated. Explanations why are even more so appreciated. Thank you, Vasto

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  • Java: Combination of recursive loops which has different FOR loop inside; Output: FOR loops indexes

    - by vvinjj
    currently recursion is fresh & difficult topic for me, however I need to use it in one of my algorithms. Here is the challenge: I need a method where I specify number of recursions (number of nested FOR loops) and number of iterations for each FOR loop. The result should show me, something simmilar to counter, however each column of counter is limited to specific number. ArrayList<Integer> specs= new ArrayList<Integer>(); specs.add(5); //for(int i=0 to 5; i++) specs.add(7); specs.add(9); specs.add(2); specs.add(8); specs.add(9); public void recursion(ArrayList<Integer> specs){ //number of nested loops will be equal to: specs.size(); //each item in specs, specifies the For loop max count e.g: //First outside loop will be: for(int i=0; i< specs.get(0); i++) //Second loop inside will be: for(int i=0; i< specs.get(1); i++) //... } The the results will be similar to outputs of this manual, nested loop: int[] i; i = new int[7]; for( i[6]=0; i[6]<5; i[6]++){ for( i[5]=0; i[5]<7; i[5]++){ for(i[4] =0; i[4]<9; i[4]++){ for(i[3] =0; i[3]<2; i[3]++){ for(i[2] =0; i[2]<8; i[2]++){ for(i[1] =0; i[1]<9; i[1]++){ //... System.out.println(i[1]+" "+i[2]+" "+i[3]+" "+i[4]+" "+i[5]+" "+i[6]); } } } } } } I already, killed 3 days on this, and still no results, was searching it in internet, however the examples are too different. Therefore, posting the programming question in internet first time in my life. Thank you in advance, you are free to change the code efficiency, I just need the same results.

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