pass-by-reference

问题 suppose I have a function which accept const reference argument pass, int func(const int &i) { /* */ } int main() { int j = 1; func(j); // pass non const argument to const reference j=2; // reassign j } this code works fine.according to C++ primer, what this argument passing to this function is like follows, int j=1; const int &i = j; in which i is a synonym(alias) of j, my question is: if i is a synonym of j, and i is defined as const, is the code: const int &i = j redelcare a non const

问题 And who has the authority to decide? Edit: Apparently I haven't succeeded in formulating my question well. I am not asking how Java's argument passing works. I know that what looks like a variable holding an object is actually a variable holding a reference to the object, and that reference is passed by value. There are lots of fine explanations of that mechanism here (in the linked threads and others) and elsewhere. The question is about the technical meaning of the term pass-by-reference.

问题 Considering that in the Ruby programming language everything is said to be an Object, I safely assumed that passing arguments to methods are done by reference . However this little example below puzzles me: $string = "String" def changer(s) s = 1 end changer($string) puts $string.class String => nil As you can see the original Object wasn't modified, I wish to know why , and also, how could I accomplish the desired behavior ie. Getting the method to actually change the object referenced by

问题 When using Objective-C I would pass a NSMutableArray from one view controller VC_A to another VC_B by simply assigning a property in VC_B as VC_B.list = self.list where self is VC_A It allows the changes done in VC_B on the list to be seen in the list in VC_A when the view controller was say popped off the navigation stack. However in Swift as arrays are passed by value, assigning as above does not work so I am stuck how to solve this. What would be the correct way to handle this now? 回答1:

问题 I'm using SharpZipLib to compress files. The library is wrapped in a plugin interface, in a separate DLL. I pass the plugin dll a ByRef parameter to keep track of the compression progress. SharpZipLib, while compressing, will periodically call a delegate sub passed when launching the compression. I can't figure out how to update the ByRef parameter when the delegate is called. If I try to assign the ByRef variable in the body of a lamba expression, I get a 'ByRef' parameter '<parametername>'

问题 Let's say I have these two functions: function fnChanger(fn) { fn = function() { sys.print('Changed!'); } } function foo() { sys.print('Unchanged'); } Now, if I call foo() , I see Unchanged , as expected. However, if I call fnChanger first, I still see Unchanged : fnChanger(foo); foo(); //Unchanged Now, I assume this is because foo is not being passed to fnChanger by reference, but I may be wrong. Why does fnChanger not change foo to print Changed! ? Furthermore, how can I get fnChanger to

问题 This: $stmt = $dbh->prepare("SELECT thing FROM table WHERE color = :color"); $stmt->bindParam(':color', $someClass->getColor()); $stmt->execute(); yields this: Runtime notice Only variables should be passed by reference though it still executes. This: $stmt = $dbh->prepare("SELECT thing FROM table WHERE color = :color"); $tempColor = $someClass->getColor(); $stmt->bindParam(':color',$tempColor); $stmt->execute(); runs without complaint. I don't understand the difference? 回答1: The second

问题 Given the example: inline string &GetLabel( ) { return m_Label; }; Where m_Label is a private class member variable. The way I think I understand it, this function will return a reference to the variable m_Label. What would be the implications of using this throughout my program and would it be a better to just return the value, instead of the reference? Thank you! 回答1: It returns a reference to the private member. There are many cases where this is desirable, but some care should be taken.

问题 Sorry to ask, its late and I can't figure a way to do it... anyone can help? $users = array( array( "name" => "John", "age" => "20" ), array( "name" => "Betty", "age" => "22" ) ); $room = array( "furniture" => array("table","bed","chair"), "objects" => array("tv","radio","book","lamp"), "users" => &$users ); var_dump $room shows: ... 'users' => & ... Which means "users" is a reference. I would like to do something like this: foreach($room as $key => $val) { if(is_reference($val)) unset($room[

问题 I created a function to print a prepared-statement-sql-string with the variables in it, based on what I found in this other StackOverflow question. Here is my code: foreach($params as $idx => $param) { if ($idx == 0) continue; $sql = str_replace('?', "'" . $param . "'", $sql, 1); } printError($sql); When I run this I get: Fatal error: Only variables can be passed by reference for line 3. However when i use $sql = preg_replace('/\?/', "'" . $param . "'", $sql, 1); for line 3 it works fine. Any