C++，通過指針傳遞泛型數組，繼承，錯誤：沒有操作符需要右手操作數

Question

我必須爲類項目實現通用二進制搜索函數。測試文件和頭文件（類定義）已經提供給我，並且不能修改。

我能夠使用我測試過的3種測試對象類型中的2種來工作，這是我難以理解的，我不知道如何進一步排除故障。

這裏是我的算法：

template <typename T, typename V> 
int binarySearch(T* list[], const V& searchValue, 
    const int firstIndex, const int lastIndex) { 

     int half = (firstIndex + lastIndex) /2; 

     // if not completly split down already 
     if(firstIndex != lastIndex && half != 0){ 

      if(searchValue < *list[half]){ 
       // lower half of array 
       binarySearch(list, searchValue, firstIndex, half); 
      } 
      else if(searchValue > *list[half]){  
       // upper half of array 
       binarySearch(list, searchValue, ++half, lastIndex); 
      } 
     } 
     else if(searchValue == *list[half]){ 
      return half; // found it 
     } 
     return -1; // didnt find it 
} 

這裏是我的3個對象的測試用例數組：

// pointers to child class objects 
Customer* customer[] = { new Customer(1002, 100000.50, "F4", "L1"), 
    new Customer(1004, 45000.90, "F1", "L3"), 
    new Customer(1003, 120000, "F3", "L2"), 
    new Customer(1001, 340000, "F2", "L4") 
}; 

// pointers to child class objects 
Employee* employee[] = { new Employee(102, 65000, "F2", "L1"), 
    new Employee(104, 45000, "F4", "L3"), 
    new Employee(103, 120000, "F1", "L2"), 
    new Employee(101, 35000, "F3", "L4") 
}; 

// pointers to parent class objects 
Person* person[] = { customer[0], 
    customer[3], 
    employee[3], 
    employee[0], 
    employee[2], 
    customer[1], 
    employee[1], 
    customer[2] 
}; 

我打電話的功能，象這樣每個對象：

// Search the customer array. -> WORKS 
cout << endl 
    << "Searching customer array for customer with cId = 1002: " 
    << (binarySearch(customer, 1002, 0, 3) != -1? "found it." : "did not find it.") 
    << endl; 

// Search the employee array. -> WORKS 
cout << "Searching employee array for employee with eId = 105: " 
    << (binarySearch(employee, 105, 0, 3) != -1? "found it." : "did not find it.") 
    << endl; 

// Search the person array. -> OPERATOR ERRORS 
cout << "Searching people array for person with name = 'Mickey Mouse': " 
    << (binarySearch(person, "Mickey Mouse", 0, 7) != -1? "found it." : "did not find it.") 
    << endl; 

搜索功能在Employee和Customer對象數組上運行良好。當試圖在Person數組上運行搜索時，我得到了每個使用的比較運算符的3個錯誤，例如：[binary'<'no操作數需要類型'Person'的右側操作數...]

我實現了操作符重載的方式，對於已經提供的函數定義中的所有三個對象，完全相同。在Person類的，我實現了以下重載運算：

bool operator ==(const Person& lhs, const Person& rhs){ 
    if(lhs.getKeyValue() == rhs.getKeyValue()) 
     return true; 
    return false; 
} 
bool operator <(const Person& lhs, const Person& rhs){ 
    if(lhs.getKeyValue() < rhs.getKeyValue()) 
     return true; 
    return false; 
} 
bool operator >(const Person& lhs, const Person& rhs){ 
    if(lhs.getKeyValue() > rhs.getKeyValue()) 
     return true; 
    return false; 
} 

在做兩個人的對象簡化測試比較，它們的比較就好了。即：

cout << "test person compare: " << ("mickey mouse" < person[1] ? "true" : "false"); 

我不知道該從哪裏拿它，方向將不勝感激。

編輯：加法（完整的人的頭文件）：

#ifndef PERSON_H 
#define PERSON_H 

#include <string> 
#include <iostream> 

using namespace std; 

namespace P03 { 
    class Person { 

    private: 
     string firstName; 
     string lastName; 

    public: 
     /* Initializes the object. 
     */ 
     Person(const string& firstName = "na", const string& lastName = "na"); 

     /* Getter methods retun the field value. 
     */ 
     string getFirstName() const; 
     string getLastName() const; 

     /* Returns the eid. 
     */ 
     string getKeyValue() const; 

     /* Returns the compound value: <lastName><space><firstName> 
     */ 
     string getName() const; 

     /* Setter methods, set the object. 
     */ 
     void setFirstName(const string& firstName); 
     void setLastName(const string& lastName); 


     /* Returns the object formatted as: 
     * Person{ firstName=<firstName>, lastName=<lastName> } 
     */ 
     virtual string toString() const; 
    }; // end Person 


    /* Displays a Person to the screen. 
    * Calls the toString() method. 
    */ 
    ostream& operator <<(ostream& out, const Person& person); 

    /* The following relational operators compare two instances of the 
    * Person class. The comparison is made on the compound value of: 
    * <lastName><space><firstName> 
    */ 
    bool operator ==(const Person& lhs, const Person& rhs); 
    bool operator !=(const Person& lhs, const Person& rhs); 
    bool operator <(const Person& lhs, const Person& rhs); 
    bool operator <=(const Person& lhs, const Person& rhs); 
    bool operator >(const Person& lhs, const Person& rhs); 
    bool operator >=(const Person& lhs, const Person& rhs); 

} // end namespace P03 

#endif 

Answer 1

你有沒有辦法把字符串轉換到一個人，所以像這樣的行失敗：

 if(searchValue < *list[half]){ 

你」如果您暫時將其更改爲：

 if (T(searchValue) < *list[half]){ 

T帽子是此代碼可以工作的唯一方式，因爲只有operator<可以採取*list[half]需要const T &在另一邊。