pass-by-value

问题 This question already has an answer here : Constructor with by-value parameter & noexcept (1 answer) Closed 1 year ago . Assume I have this function void foo() noexcept { // Safely noexcept code. } And then this class: class Bar { Bar(const Bar&) { ... } // Is not noexcept, so might throw // Non movable: Bar(Bar&&) = delete; }; Now, I need to modify foo() to receive a Bar by value: void foo(Bar bar) // noexcept? { // Safely noexcept code } I assume the copy of Bar is done before the call to

问题 As much as I enjoy C++ programming, there is one thing I really don't get. To me, it seems that the most common way of programming a function is like this: some_function(a variable) do something according to the data in the variable example: bool match_name(const std::string& name) { return this->name == name; } I find myself using const ref for 90% of all function parameters in my code (maybe I'm doing something wrong). My question is: Why is a copy of the variable the "default" type of

问题 As much as I enjoy C++ programming, there is one thing I really don't get. To me, it seems that the most common way of programming a function is like this: some_function(a variable) do something according to the data in the variable example: bool match_name(const std::string& name) { return this->name == name; } I find myself using const ref for 90% of all function parameters in my code (maybe I'm doing something wrong). My question is: Why is a copy of the variable the "default" type of

问题 In C++ where we use pass by reference we reference the address of whatever it is that we passed from the argument to the parameter of the function which is essentially a pointer right? So while they are essentially the same thing, alias and all, doesnt a pointer require memory space as well? So shouldnt whatever we have in a parameter function let us call B point to the memory location of whatever the argument was that was passed let us call A which in turn is the memory location of our value

问题 In C++ where we use pass by reference we reference the address of whatever it is that we passed from the argument to the parameter of the function which is essentially a pointer right? So while they are essentially the same thing, alias and all, doesnt a pointer require memory space as well? So shouldnt whatever we have in a parameter function let us call B point to the memory location of whatever the argument was that was passed let us call A which in turn is the memory location of our value

问题 I am studying/learning the bahavior of ByVal and ByRef when it comed to working with a call object. So I created this class PersonModel Private Type TPerson firstName As String lastName As String End Type Private this As TPerson Public Property Get firstName() As String firstName = this.firstName End Property Public Property Let firstName(ByVal strNewValue As String) this.firstName = strNewValue End Property Public Property Get lastName() As String lastName = this.lastName End Property Public

问题 Is it possible in Java (or Groovy) to pass an object by reference into a wrapper object (ie: List or Map)? Example code (in Groovy): def object = null def map = [object: object] object = new Object() Unfortunately map.object remains null even though the object variable doesn't, so obviously the original creation of the map was done by value not by reference. Is it possible to create a List or Map via references so that when the object outside the List or Map changes the change is reflected

问题 I have this sample code: public class MyClass { public int Value { get; set; } } class Program { public static void Foo(MyClass v) { v.Value = 2; v = new MyClass(); v.Value = 3; } static void Main(string[] args) { var m = new MyClass(); m.Value = 1; Foo(m); Console.Write(m.Value); Console.ReadLine(); } } I would like to understand why the output is 2 and not 3, could you please give me some clear explanation? Thanks 回答1: I will go through with you, step by step via the debugger and we will

问题 I sent a reference to a bool object, and I modified it within a method. After the method finished it's execution, the value of the bool outside the method was unchanged. This leads me to believe that Python's bools are passed by value. Is that true? What other Python types behave that way? 回答1: Python variables are not "references" in the C++ sense. Rather, they are simply local names bound to an object at some arbitrary location in memory. If that object is itself mutable, changes to it will

问题 I'm trying to declare a pointer and pass that pointer to a function where memory is allocated. Here is a minimal example: #include <string> #include <iostream> using namespace std; void alloc_mem(int &size, double *x); int main() { double *X; int imax; alloc_mem(imax, X); cout << "imax = " << imax << endl; for (int i = 0; i < imax; i++) { cout << "X = " << X[i] << endl; } delete[]X; return 0; } void alloc_mem(int &size, double *x) { size = 10; x = new double[size]; for (int i = 0; i < size; i