return-by-reference

问题 How can I pass a variable by reference in scheme? An example of the functionality I want: (define foo (lambda (&x) (set! x 5))) (define y 2) (foo y) (display y) ;outputs: 5 Also, is there a way to return by reference? 回答1: See http://community.schemewiki.org/?scheme-faq-language question "Is there a way to emulate call-by-reference?". In general I think that fights against scheme's functional nature so probably there is a better way to structure the program to make it more scheme-like. 回答2:

问题 Whew, that was a long title. Here's my problem. I've got a template class in C++ and I'm overloading the [] operator. I have both a const and a non-const version, with the non-const version returning by reference so that items in the class can be changed as so: myobject[1] = myvalue; This all works until I use a boolean as the template parameter. Here's a full example that shows the error: #include <string> #include <vector> using namespace std; template <class T> class MyClass { private:

问题 Whew, that was a long title. Here's my problem. I've got a template class in C++ and I'm overloading the [] operator. I have both a const and a non-const version, with the non-const version returning by reference so that items in the class can be changed as so: myobject[1] = myvalue; This all works until I use a boolean as the template parameter. Here's a full example that shows the error: #include <string> #include <vector> using namespace std; template <class T> class MyClass { private:

问题 I am reasoning about the best approach to return references to objects created inside a method, like in the following situation: class A{ public: A(){} ~A(){} }; class Foo{ public: Foo(){} ~Foo(){} A& create(int random_arg){ // create object A and return its reference } }; void other_method(){ Foo f; A a = f.create(); // do stuff with a { I have considered three possible solutions: create a raw pointer and return a reference, but this is bad because there is no guarantee that the object will

问题 Can you explain to me the difference between returning value, reference to value, and const reference to value? Value: Vector2D operator += (const Vector2D& vector) { this->x += vector.x; this->y += vector.y; return *this; } Not-const reference: Vector2D& operator += (const Vector2D& vector) { this->x += vector.x; this->y += vector.y; return *this; } Const reference: const Vector2D& operator += (const Vector2D& vector) { this->x += vector.x; this->y += vector.y; return *this; } What is the

问题 Can you explain to me the difference between returning value, reference to value, and const reference to value? Value: Vector2D operator += (const Vector2D& vector) { this->x += vector.x; this->y += vector.y; return *this; } Not-const reference: Vector2D& operator += (const Vector2D& vector) { this->x += vector.x; this->y += vector.y; return *this; } Const reference: const Vector2D& operator += (const Vector2D& vector) { this->x += vector.x; this->y += vector.y; return *this; } What is the

问题 In the old days, if I wanted a string representation of an object A , I would write something with the signature void to_string(const A& a, string& out) to avoid extra copies. Is this still the best practice in C++11, with move semantics and all? I have read several comments on other contexts that suggest relying on RVO and instead writing string to_string(const A& a) . But RVO is not guaranteed to happen! So, how can I, as the programmer of to_string, guarantee the string is not copied

问题 In the old days, if I wanted a string representation of an object A , I would write something with the signature void to_string(const A& a, string& out) to avoid extra copies. Is this still the best practice in C++11, with move semantics and all? I have read several comments on other contexts that suggest relying on RVO and instead writing string to_string(const A& a) . But RVO is not guaranteed to happen! So, how can I, as the programmer of to_string, guarantee the string is not copied

问题 The following code invokes undefined behaviour. int& foo() { int bar = 1234; return bar; } g++ issues a warning: warning: reference to local variable ‘bar’ returned [-Wreturn-local-addr] clang++ too: warning: reference to stack memory associated with local variable 'bar' returned [-Wreturn-stack-address] Why is this not a compile error (ignoring -Werror )? Is there a case where returning a ref to a local var is valid? EDIT As pointed out, the spec mandates this be compilable. So, why does the

问题 I'm confused about how PHP variable references work. In the examples below, I want to be able to access the string hello either as $bar[0] or $barstack[0][0] . It would seem that passing the array by reference in step 1 should be sufficient. The second example does not work. $foostack[0]0] is the string hello , but $foo[0] doesn't exist. At some point, the first element of $foostack becomes a copy of $foo, instead of a reference. The problem lies in the first line of step 2: When I push a