Question

我有一个链接器错误，我似乎没弄明白：

默认构造函数将name设置为“Unknown”，将Office Number设置为nextOfficeNo，将Employee number设置为nextEmpId，将Department number设置为0，将Employee Position设置为Entry，将经验年份设置为0，将salary设置为0。确保所有静态属性的值增加1.

第二个构造函数根据传递给函数的内容设置属性。 Employee Salary的值仍将设置为0，Office Number和Employee number的值分别设置为nextOfficeNo和nextEmpId。同样，构造函数应该将所有静态属性的值递增1。也; totalNumOfEmployees的值必须在创建任何对象之前初始化为0，在创建Employee类的每个对象时（在每个构造函数中）递增，并在Employee类的对象超出作用域时（在析构函数中）递减。

在创建任何对象之前，必须将nextEmpId的值初始化为1000，并且必须在每个构造函数中创建类Employee的每个对象时递增。

在创建任何对象之前，必须将nextOfficeNo的值初始化为10，并且必须在每个构造函数中创建类Employee的每个对象时递增。

这是我的标题类：

#include <iostream>
#include <string>
using namespace std;

class Employee
{
private:
    string name;
    const long officeNo;
    const long empId;
    int deptNo;
    char empPosition; // ‘E’: entry level, ‘M’: manager, ‘D’: Director, ‘P’:Project_leader
    int yearOfExp;
    float salary;
    static int totalNumofEmployees;
    static int nextEmpId;
    static int nextOfficeNo;
public:
    Employee();
    ~Employee();
    Employee(string theName, int theDeptNo, char theEmpPosition, int theYearOfExp);
    void Print() const ;
    void GetInfo();
    friend void setSalary(Employee& );
 };

这是我的CPP班：
我的构造函数中存在问题：

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

Employee::Employee()
  : officeNo(nextOfficeNo), empId(nextEmpId)
{
    name = "Unknown";
    deptNo = 0;
    empPosition = 'E';
    yearOfExp = 0;
    salary = 0;
    totalNumofEmployees = 0;
    nextEmpId = 1000;
    nextOfficeNo = 10;
    totalNumofEmployees++;
    nextEmpId++;
    nextOfficeNo++;
}

Employee::Employee(string theName, int theDeptNo, char theEmpPosition, int theYearOfExp)
  : officeNo(nextOfficeNo), empId(nextEmpId)
{
    name = theName;
    deptNo = theDeptNo;
    empPosition = theEmpPosition;
    yearOfExp = theYearOfExp;
    salary = 0;
    totalNumofEmployees = 0;
    nextEmpId = 1000;
    nextOfficeNo = 10;
    totalNumofEmployees++;
    nextEmpId++;
    nextOfficeNo++;
}

这里是错误：

{Undefined symbols for architecture x86_64:
"Employee::nextOfficeNo", referenced from:
Employee::Employee() in Employee.o
Employee::Employee(std::__1::basic_string<char, std::__1::char_traits<char>, std::__1::allocator<char> >, int, char, int) in Employee.o
"Employee::totalNumofEmployees", referenced from:
Employee::Employee() in Employee.o
Employee::Employee(std::__1::basic_string<char, std::__1::char_traits<char>, std::__1::allocator<char> >, int, char, int) in Employee.o
Employee::~Employee() in Employee.o
Employee::Print() const in Employee.o
"Employee::nextEmpId", referenced from:
Employee::Employee() in Employee.o
Employee::Employee(std::__1::basic_string<char, std::__1::char_traits<char>,  std::__1::allocator<char> >, int, char, int) in Employee.o
ld: symbol(s) not found for architecture x86_64
clang: error: linker command failed with exit code 1 (use -v to see invocation)
}

Answer 1

您已声明静态成员变量，但您忘记定义它们：

§9.4.2/ 2 静态数据成员在其类定义中的声明不是定义，可能是除了cv-qualified void之外的不完整类型。静态数据成员的定义应出现在命名空间范围包含成员的类定义。在命名空间范围的定义中，名称静态数据成员应使用::运算符通过其类名限定。初始化程序静态数据成员定义中的表达式属于其类的范围。

// Example:
class process {
    static process* run_chain;
    static process* running;
};
process* process::running = get_main();
process* process::run_chain = running;

在你的情况下：

// add this to your .cpp
int Employee::totalNumofEmployees = 0;
int Employee::nextEmpId = 1000;
int Employee::nextOfficeNo = 10;

从构造函数中删除这些赋值：

totalNumofEmployees = 0;
nextEmpId = 1000;
nextOfficeNo = 10;

否则，无论何时创建对象，都会重置这些值。

Apple mach O链接器

1 个答案: