In your expression BinaryNode * & t) BinaryNode* & t ————- —– BinaryNode pointer t is reference variable so t is reference to pointer of BinaryNode class. Pointer of the address of t? You are confused ampersand & operator in c++. that give address of an variable. but syntax is different. ampersand & in front … Read more
So you can write code like this: void f( const string & s ) { } f( “foobar” ); Although strictly speaking what is actually happening here is not the literal being bound to a const reference – instead a temprary string object is created: string( “foobar” ); and this nameless string is bound to … Read more
What is the difference between ptr and rptr? If you do char *world = “World”; rptr = world; and then print str, it will print “World”. If you do ptr = world; and then print str, it will print “Hello”.
I just experienced this same problem, calling bind_param via call_user_func_array and passing an array of parameters. The solution is to modify the values in the array to be referenced. It’s not elegant but it works. call_user_func_array(array($stmt, ‘bind_param’), makeValuesReferenced($passArray)); function makeValuesReferenced($arr){ $refs = array(); foreach($arr as $key => $value) $refs[$key] = &$arr[$key]; return $refs; }
If you know what you’re doing, you can make a vector of references using std::reference_wrapper: #include <functional> #include <vector> int main() { std::vector<std::reference_wrapper<int>> iv; int a = 12; iv.push_back(a); // or std::ref(a) // now iv = { 12 }; a = 13; // now iv = { 13 }; } Note of course that any … Read more
The answer to this, as it turns out, is Yes. However it is not a tidy syntax as it uses a sort of sub-statement, and leaves the current scope littered with an extra reference variable. Consider the following code: <?php $array = array( // Creates Key1 and assigns the value to it // A copy … Read more
A reference is semantically the following: T& <=> *(T * const) const T& <=> *(T const * const) T&& <=> [no C equivalent] (C++11) As with other answers, the following from the C++ FAQ is the one-line answer: references when possible, pointers when needed. An advantage over pointers is that you need explicit casting in … Read more
There are several reasons why returning references (or pointers) to the internals of a class are bad. Starting with (what I consider to be) the most important: Encapsulation is breached: you leak an implementation detail, which means that you can no longer alter your class internals as you wish. If you decided not to store … Read more
That is taking the parameter by reference. So in the first case you are taking a pointer parameter by reference so whatever modification you do to the value of the pointer is reflected outside the function. Second is the simlilar to first one with the only difference being that it is a double pointer. See … Read more
You don’t have to create the default vector if you don’t use it. You just have to ensure the declaration is done outside the if block. fn accept(input: &Vec<String>) { let def; let vec = if input.is_empty() { def = vec![“empty”.to_string()]; &def } else { input }; // … do something with `vec` } Note … Read more