c++ - C++:如何返回集合的迭代器指向的值？

Question

我从集合上的迭代器返回的值给出了内存位置而不是值。如何访问迭代器指向的元素？

我使用了类似的迭代器循环，在此之前效果很好，因为返回了迭代器指向的值(对于双端队列和 vector 模板类)。

这个问题是我今天早些时候遇到困难的脚本(C++ : adding an object to a set)的后续。

现在，脚本如下所示:

头文件

#ifndef EMPLOYEE_HH
#define EMPLOYEE_HH

#include <set>
#include <string>
#include <iostream>

using namespace std ;

class Employee {

public:
  // Constructor
  Employee(const char* name, double salary) :
    _name(name),
    _salary(salary) {
  }

  // Accessors
  const char* name() const { 
    return _name.c_str() ;
  }

  double salary() const {
    return _salary ;
  }

  // Print functions
  void businessCard(ostream& os = cout) const {
    os << "   +--------------------+  " << endl
       << "   |  ACME Corporation  |  " << endl 
       << "   +--------------------+  " << endl
       << "   Name: " << name() << endl
       << "   Salary: " << salary() << endl ;
  }

private:
  string _name ;
  double _salary ;

} ;

class Manager : public Employee {

public:
  //Constructor
  Manager(const char* _name, double _salary):
    Employee(_name, _salary),
    _subordinates() {
  }

  // Accessors & modifiers
  void addSubordinate(Employee& empl) {
    _subordinates.insert(&empl);
  }

  const set<Employee*>& listOfSubordinates() const {
    return _subordinates;
  }

  void businessCard(ostream& os = cout) const {
    Employee::businessCard() ;
    os << "   Function: Manager" << endl ;

    set<Employee*>::iterator iter ;
    iter = _subordinates.begin() ;

    os << "   Subordinates:" << endl ;
    if(_subordinates.empty()==true) {
      os << "      Nobody" << endl;
    }
    while(iter!=_subordinates.end()) {
      os << "      " << *iter << endl ;  // <-- returns the memory location 
      ++iter ;
    }

  }

private:
  set<Employee*> _subordinates ;
} ;

#endif

主脚本

#include <string>
#include <iostream>
#include "Employee.hh"

using namespace std ;

int main() {

   Employee emp1("David", 10000) ;
   Employee emp2("Ivo", 9000) ;
   Manager mgr1("Oscar", 18000) ;   // Manager of Ivo and David
   Manager mgr2("Jo", 14000) ;
   Manager mgr3("Frank", 22000) ;  // Manager of Jo and Oscar (and Ivo and David)

   mgr1.addSubordinate(emp1) ;
   mgr1.addSubordinate(emp2) ;
   mgr3.addSubordinate(mgr1) ;
   mgr3.addSubordinate(mgr2) ;

   cout << '\n' ;
   emp1.businessCard() ;

   cout << '\n' ;
   emp2.businessCard() ;

   cout << '\n' ;
   mgr1.businessCard() ;

   cout << '\n' ;
   mgr2.businessCard() ;

   cout << '\n' ;
   mgr3.businessCard() ;

   cout << '\n' ;       
   return 0;
}

任何帮助深表感谢。

Answer 1

如果此地址:*iter是地址，则此地址:*(*iter)是*iter指向的值。

这应该在您的情况下有效:

while(iter != _subordinates.end()) 
{
   os << "      " << **iter << endl ;  // <-- returns the value which is the set
   ++iter;
}

编辑:此问题已解决: (*iter)->name()