recursion

问题 the following function creates new folder on my server via xmlrpc var createFolder = function(folder_name) { var defer = Q.defer(); client.methodCall('create_folder', [sessionID, folder_name], function(err, resp) { if (err) { if (err.responseString && err.responseString.match('already exist')) { //call the same function recursively with folder_name+Math.round(Math.random()*100) } else { defer.reject(err); } } else { defer.resolve(folder_name); } }); return defer.promise; } The functions

问题 the following function creates new folder on my server via xmlrpc var createFolder = function(folder_name) { var defer = Q.defer(); client.methodCall('create_folder', [sessionID, folder_name], function(err, resp) { if (err) { if (err.responseString && err.responseString.match('already exist')) { //call the same function recursively with folder_name+Math.round(Math.random()*100) } else { defer.reject(err); } } else { defer.resolve(folder_name); } }); return defer.promise; } The functions

问题 I have a class that contains some fields. I need to compare instances of this class by value, so I defined GetHashCode and Equals accordingly. Because the class allows circular references, I need a mechanism to avoid infinite recursion (for a more detailed explanation see Value-equals and circular references: how to resolve infinite recursion?). I solved this problem by modifying my Equals method so that it keeps track of the comparisons done before: class Foo { public string Name { get; set;

问题 I have a class that contains some fields. I need to compare instances of this class by value, so I defined GetHashCode and Equals accordingly. Because the class allows circular references, I need a mechanism to avoid infinite recursion (for a more detailed explanation see Value-equals and circular references: how to resolve infinite recursion?). I solved this problem by modifying my Equals method so that it keeps track of the comparisons done before: class Foo { public string Name { get; set;

问题 I'm trying to wrap my mind around xslt. A number of questions here on stackoverflow help ( XSLT templates and recursion and XSLT for-each loop, filter based on variable ) but I'm still kinda puzzled. I guess I'm "thinking of template as functions" ( https://stackoverflow.com/questions/506348/how-do-i-know-my-xsl-is-efficient-and-beautiful ) Anyway...my data is <Entities> <Entity ID="8" SortValue="0" Name="test" ParentID="0" /> <Entity ID="14" SortValue="2" Name="test2" ParentID="8" /> <Entity

问题 Say there is any given list in Scheme. This list is ‘(2 3 4) I want to find all possible partitions of this list. This means a partition where a list is separated into two subsets such that every element of the list must be in one or the other subsets but not both, and no element can be left out of a split. So, given the list ‘(2 3 4) , I want to find all such possible partitions. These partitions would be the following: {2, 3} and {4}, {2, 4} and {3}, and the final possible partition being

问题 As it currently stands, this question is not a good fit for our Q&A format. We expect answers to be supported by facts, references, or expertise, but this question will likely solicit debate, arguments, polling, or extended discussion. If you feel that this question can be improved and possibly reopened, visit the help center for guidance. Closed 6 years ago . I have got a $branch object that can contain other $branch objects: $branch->children(); Each of them can have $apples as children.

问题 I am starting on learning Prolog. This program tries to get all occurrences of a given element: occurences(_, [], Res):- Res is []. occurences(X, [X|T], Res):- occurences(X,T,TMP), Res is [X,TMP]. occurences(X, [_|T], Res):- occurences(X,T,Res). But here is the error: ?- occurences(a,[a,b,c,a],Res). ERROR: is/2: Arithmetic: `[]/0' is not a function ^ Exception: (11) _G525 is [] ? creep Exception: (10) occurences(a, [], _G524) ? creep Exception: (9) occurences(a, [a], _G524) ? creep Exception:

问题 I'm reading The Little Schemer and feel confused about the following code: ((lambda (len) (lambda (l) (cond ((null? l) 0) (else (+ 1 (len (cdr l))))))) eternity) (define eternity (lambda (x) (eternity x))) The code is to determine the empty list, otherwise it never stops. Why is the " len " not recursion? 回答1: Although it can be dangerous to apply textual substitution to Lisp forms (since there are dangers of multiple evaluation, etc.), in this case it may help to look at this form and see

问题 I'm reading The Little Schemer and feel confused about the following code: ((lambda (len) (lambda (l) (cond ((null? l) 0) (else (+ 1 (len (cdr l))))))) eternity) (define eternity (lambda (x) (eternity x))) The code is to determine the empty list, otherwise it never stops. Why is the " len " not recursion? 回答1: Although it can be dangerous to apply textual substitution to Lisp forms (since there are dangers of multiple evaluation, etc.), in this case it may help to look at this form and see