如何正确初始化副本构造函数（以类为参考的构造函数）

Question

如何初始化具有引用的类的构造函数的副本构造函数。我只是不知道在冒号后面放什么来初始化它。

class Me{
 public:   
    Me (const otherMe& t)
    :other_me(t)
    {}
    //copy constructor
    Me(const Me& me)
    : /*what do you put here in order to write 
    the line of code bellow. I tried t(t), and gives me the 
    warning 'Me::t is initialized with itself [-Winit-self]' */

    {cout << t.getSomthing() << endl;}

 private:
    const otherMe& other_me;
};

Answer 1

假设您有两个课程，void SomeMethod<T>(List<T> oldRecords, List<T> newRecords, List<string> propertiesOfT) { // get the list of property to match var properties = propertiesOfT.Select(prop => typeof(T).GetProperty(prop)).ToList(); // Get all old record where we don't find any matching new record where all the property equal that old record var notMatch = oldRecords.Where(o => !newRecords.Any(n => properties.All(prop => prop.GetValue(o).Equals(prop.GetValue(n))))).ToList(); } 和public class test { public int id { get; set; } = 0; public string desc { get; set; } = ""; public test(string s, int i) { desc = s;id = i; } } private void Main() { var oldRecords = new List<test>() { new test("test",1), new test("test",2) }; var newRecords = new List<test>() { new test("test1",1), new test("test",2) }; SomeMethod(oldRecords, newRecords, new List<string>() { "id", "desc" }); } ：

Value

我们可以这样编写构造函数和复制构造函数：

Wrapper

如果class Value { // stuff... }; class Wrapper; // This one contains the reference 是一个const引用，此方法也可以使用！此外，如果您可以将class Wrapper { Value& val; public: Wrapper(Value& v) : val(v) {} Wrapper(Wrapper const& w) : val(w.val) {} }; 汇总编写，它将自动获得一个副本构造函数：

Value&