将const对象显式传递给构造函数,该构造函数将const引用到多态类

Question

我遇到了类的问题,将一个const对象(多态结构)传递给一个显式构造函数,该构造函数对该多态结构的基类进行了const引用.这是示例(这不是我的代码,这里是解释)

class Base 
{
...
}

class Derived:public Base
{
...
}

class Problem 
{
    Problem(const Base&);
...
}

void myFunction(const Problem& problem) 
{
    ...
}

int main() 
{
    //explicit constructor with non const object
    Derived d;
    Problem no1(d); //this is working fine 
    myFunction(no1);

    //implicit constructor with const object
    Problem no2=Derived(); //this is working fine, debugged and everything called fine
    myFunction(no2);   //is working fine

    //explicit constructor with const object NOT WORKING
    Problem no3(Derived());  //debugger jumps over this line (no compiler error here)
    myFunction(no3);   //this line is NOT COMPILING at all it says that:
    //no matching function for call to myFunction(Problem (&)(Derived))
    //note: candidates are: void MyFunction(const Problem&)
}

只有当我将Derived对象显式地转换为它的基类Base时,似乎它与第二个版本(显式构造函数调用问题)一起工作正常:

Problem(*(Base*)&Derived);

我没有意识到有问题和明确地调用Problem类的构造函数之间的区别.谢谢!

Answer 1

问题是你没有声明一个对象,而是一个函数:

Problem no3(Derived());
// equivalent to:
Problem no3(Derived); // with parameter name omitted

使用:

Problem no3((Derived()));
// extra parens prevent function-declaration interpretation
// which is otherwise required by the standard (so that the code isn't ambiguous)

这是C++继承的C语句声明的一个怪癖.

更多例子:

void f(int(a)); /* same as: */ void f(int a);

void g() {
  void function(int);    // declare function
  void function(int());  // we can declare it again
  void function(int=42); // add default value
  function();            // calls ::function(42) ('function' in the global scope)
}
// 'function' not available here (not declared)

void function(int) {} // definition for declarations inside g above