C++,通过指针传递泛型数组,继承,错误:没有操作符需要右手操作数
我必须为类项目实现通用二进制搜索函数。测试文件和头文件(类定义)已经提供给我,并且不能修改。C++,通过指针传递泛型数组,继承,错误:没有操作符需要右手操作数
我能够使用我测试过的3种测试对象类型中的2种来工作,这是我难以理解的,我不知道如何进一步排除故障。
这里是我的算法:
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
}
这里是我的3个对象的测试用例数组:
// 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]
};
我打电话的功能,象这样每个对象:
// 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对象数组上运行良好。当试图在Person数组上运行搜索时,我得到了每个使用的比较运算符的3个错误,例如:[binary'<'no操作数需要类型'Person'的右侧操作数...]
我实现了操作符重载的方式,对于已经提供的函数定义中的所有三个对象,完全相同。在Person类的,我实现了以下重载运算:
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;
}
在做两个人的对象简化测试比较,它们的比较就好了。即:
cout << "test person compare: " << ("mickey mouse" < person[1] ? "true" : "false");
我不知道该从哪里拿它,方向将不胜感激。
编辑:加法(完整的人的头文件):
#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
你有没有办法把字符串转换到一个人,所以像这样的行失败:
if(searchValue < *list[half]){
你”如果您暂时将其更改为:
if (T(searchValue) < *list[half]){
T帽子是此代码可以工作的唯一方式,因为只有operator<
可以采取*list[half]
需要const T &
在另一边。
你真棒的家伙,谢谢。不知道我明白为什么其他两个班没有它,但他们现在都工作。愿施瓦茨与你同在。 – SomeRandomDeveloper 2013-03-12 02:19:13
对不起,他们编译得出结论,但实际上并没有做比较的权利,但重新阅读您的文章让我意识到您的意思是为了帮助我进行调试,谢谢,它帮助我查看对象现在用调试器。 – SomeRandomDeveloper 2013-03-12 13:31:03
您没有将'Person'对象传递给搜索,您正在传递一个字符串。你的比较函数(如((searchValue Joe 2013-03-12 01:45:17
这些是我的想法,但另外两个工作!我很困惑。我认为他们现在都需要下面的解决方案,我认为它的工作。 – SomeRandomDeveloper 2013-03-12 02:20:48
你可以显示头文件吗? – 2013-03-12 02:21:56