c++ - 未在此范围内声明

Question

http://pastebin.com/4gvcQm7P

#include <iostream>

using namespace std;

int GenerateID()
{
    static int nextID = 0;
    return nextID++;
}

void PrintInformation(Employee EmployeeName)
{
    cout << EmployeeName << "'s ID is: " << EmployeeName.ID << endl;
    cout << EmployeeName << "'s age is: " << EmployeeName.age << endl;
    cout << EmployeeName << "'s wage is: " << EmployeeName.wage << endl;
}

int main()
{

    struct Employee
    {
        int ID;
        int age;
        float wage;
    };

    Employee Dominic;
    Employee Jeffrey;

    Dominic.ID = GenerateID();
    Dominic.age = 22;
    Dominic.wage = 7.10;

    Jeffrey.ID = GenerateID();
    Jeffrey.age = 28;
    Dominic.wage = 7.10;

    PrintInformation(Dominic);
    PrintInformation(Jeffrey);

    return 0;
}

/*
C:\CBProjects\Practise\main.cpp|11|error: variable or field 'PrintInformation' declared void|
C:\CBProjects\Practise\main.cpp|11|error: 'Employee' was not declared in this scope|
C:\CBProjects\Practise\main.cpp||In function 'int main()':|
C:\CBProjects\Practise\main.cpp|39|error: 'PrintInformation' was not declared in this scope|
||=== Build finished: 3 errors, 0 warnings (0 minutes, 0 seconds) ===|
*/

上面的 pastebin 链接显示了我使用的代码和构建报告。根据这份报告，我试图转发声明结构而不包括成员，然后出现“不完整类型”错误。

解决方案是什么？

编辑:我正在使用 c++11

编辑 2:如果我尝试转发声明结构，包括成员，会发生以下情况:

http://pastebin.com/rrt4Yjes#

Answer 1

有两种解决方案:使 Employee 成为非本地类/结构或使 PrintInformation 成为模板。对于第一个解决方案，只需移动 Employee before PrintInformation。第二种解决方案是:

template< typename Employee >
void PrintInformation(const Employee& EmployeeName)
{
    cout << " EmployeeName's ID is: " << EmployeeName.ID << endl;
    cout << " EmployeeName's age is: " << EmployeeName.age << endl;
    cout << " EmployeeName's wage is: " << EmployeeName.wage << endl;
}

请注意，在任何情况下，您都不希望 Employee 的拷贝只是为了打印一些信息，因此请将 PrintInformation 的参数设为常量引用，如上所示。