Question

我很抱歉这个问题，因为它可能会在这里回答，但到目前为止我的搜索都没有结果。

如果我使用参数化构造函数，我可以将我的类对象传递给输出函数，一切都很好。如果我使用默认构造函数，它将失败：

1>c:\<path>\project_04.cpp(152): error C2664: 'printCheck' : cannot convert parameter 1 from 'AdamsEmployee (__cdecl *)(void)' to 'AdamsEmployee'
1>          No constructor could take the source type, or constructor overload resolution was ambiguous

当我尝试将其传递给输出函数时，鼠标悬停在我的对象上：

Error: No suitable constructor exists to convert from "AdamsEmployee ()" to "AdamsEmployee"

这是我的默认构造函数：

AdamsEmployee::AdamsEmployee()
{
    AdamsEmployee::employeeNumber = -1;
    AdamsEmployee::employeeName = "";
    AdamsEmployee::employeeAddress = "";
    AdamsEmployee::employeePhone = "";
    AdamsEmployee::employeeHourlyWage = 0.0;
    AdamsEmployee::employeeHoursWorked = 0.0;
}

这是我的参数化构造函数：

AdamsEmployee::AdamsEmployee(int employeeNumber, string employeeName, string
employeeAddress, string employeePhone, double employeeHourlyWage,
doubleemployeeHoursWorked )
{
    AdamsEmployee::employeeNumber = employeeNumber;
    AdamsEmployee::employeeName = employeeName;
    AdamsEmployee::employeeAddress = employeeAddress;
    AdamsEmployee::employeePhone = employeePhone;
    AdamsEmployee::employeeHourlyWage = employeeHourlyWage;
    AdamsEmployee::employeeHoursWorked = employeeHoursWorked;
}

调用输出的行：

printCheck( emp1 );

输出功能：

void printCheck( AdamsEmployee employee )
{
// Display the mock paycheck.
cout << "----------------------------------H&H Systems----------------------------------" << endl;
cout << "\nPay to the order of " << employee.getName() << ".....$" << employee.calcPay() << endl;
// Display the simulated paystub.
cout << "\nGoliath National Bank" << endl;
cout << "-------------------------------------------------------------------------------" << endl;
cout << "Hours worked: " << employee.getHoursWorked() << endl;
cout << "Hourly wage: " << employee.getWage() << endl;
} // End printCheck()

如果我添加参数，一切正常。搜索返回了许多似乎不适用的情况。您还需要更多信息吗？

我做错了什么？

编辑：感谢您的帮助！

Answer 1

您的错误表示您正在传递函数，就好像您已声明AdamsEmployee emp1()一样。正如一篇评论所提到的，这可能是由于解析模糊性。它是如此常见，它有一个完整的stackoverflow标记：https://stackoverflow.com/questions/tagged/most-vexing-parse

Answer 2

您可能缺少默认构造函数的类中构造函数的声明。

C ++“没有合适的构造函数可以从<default constructor =”“>转换为参数化构造函数</default>

2 个答案: