c++ - 显式地将 const 对象传递给构造函数，该构造函数采用对多态类的 const 引用

Question

我的类遇到了问题，将 const 对象(多态结构)传递给显式构造函数，该构造函数采用对该多态结构的基类的 const 引用。这是示例(这不是我的代码，这里是为了解释)

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 对象显式转换为其基类 Base 时，它​​似乎在第二个版本(对问题的显式构造函数调用)上工作良好:

Problem(*(Base*)&Derived);

我没有意识到隐式调用和显式调用 Problem 类的构造函数之间的区别。谢谢!

Answer 1

问题是你不是在声明一个对象，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 声明语法的怪癖。

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