Question

我在显示我的子类中的一些字符串时遇到问题。我正在尝试使用函数，但我不确定为什么我没有得到这些字符串的内容。

class Employee{
    string FN, LN, JT;
    double Income;

public:
    char const *getters(){
        return FN.data(), LN.data(), JT.data(); //=========>getting the content of strings
    }
    virtual char const *getAccess()=0;
    Employee(char const *fn, char const *ln, char const *jt, double inc){

        if(fn==0) throw Exception(1, "Sorry, First Name is Null");
        if(ln==0) throw Exception(2, "Sorry, Last Name is Null");
        if(jt==0) throw Exception(3, "Sorry Job Title is Null");
        if(inc<=0) throw Exception(4, "Sorry, The Income is Null");

        FN=fn;
        LN=ln;
        JT=jt;
        Income=inc;
    }
};

class Programmer: public Employee{
public:
    Programmer(char const *fn, char const *ln, double inc):
        Employee(fn,ln,"Programmer", inc)
    {}
    char const *getAccess(){
        return "You have access to Meeting Room + Development Office";
    }
};

//=========The Main============
int main(){
    Employee *acc[3];

    try{
        acc[0]=new Programmer("Juan", "Villalobos", 60000);
        acc[1]=new Director("Jorge", "Villabuena", 70000);
        acc[2]=new ProdSupport("Pedro", "Villasmil", 80000);
        for(int i=0; i<3; i++){
            cout << acc[i]->getters() << endl;    //=============>Displaying the strings
            cout << acc[i]->getAccess() << endl;
        }
    } catch(Exception acc){
        cout << "Err:" << acc.getErrCode() << " Mess:" << acc.getErrMess() << endl;
    }

    return 0;
}

所以，我猜我的功能没有做我想要的，显示名字和姓氏。我做错了什么？

Answer 1

,逗号运算符的结果是右手值，因此return FN.data(), LN.data(), JT.data();实际上与return JT.data();相同，这就是您所看到的。

要做你正在尝试的事情，请尝试这样做：

std::vector<std::string> getValues() const {
    std::vector<std::string> arr(3);
    arr.push_back(FN);
    arr.push_back(LN);
    arr.push_back(JT);
    return arr;
}

std::vector<std::string> arr = acc[i]->getValues();
for (std::vector<std::string>::const_iterator iter = arr.begin(), end = arr.end(); iter != end; ++iter) {
    cout << *iter << " ";
}
cout << endl;

或者，将cout逻辑移入类本身：

void displayValues() const {
    cout << FN << " " << LN << " " << JT << endl;
    cout << getAccess() << endl;
}

for(int i=0; i<3; i++){
    acc[i]->displayValues();
}

Answer 2

我不认为混合char*和string。比较晚。

这会编译

char const *getters(){
    return FN.data(), LN.data(), JT.data();
}

但你可能想要的是

char const *getters(){
    return (FN + LN + JT).data();
}

我会像这样重写您的程序：

class Employee{
    string FN, LN, JT;
    double Income;

public:
    string getters(){
        return FN + " " + LN + " " + JT;
    }

    virtual string getAccess()=0;

    Employee(string const &fn, string const &ln, string const &jt, double inc) :
        FN(fn), LN(ln), JT(jt), Income(inc)
    {
    }
};

class Programmer: public Employee{
public:
    Programmer(string const &fn, string const &ln, double inc):
        Employee(fn,ln,"Programmer", inc)
    {}

    string getAccess(){
        return "You have access to Meeting Room + Development Office";
    }
};

//=========The Main============
int main()
{
    std::vector<Employee> acc;

    acc.push_back(Programmer("Juan", "Villalobos", 60000));
    acc.push_back(Director("Jorge", "Villabuena", 70000));
    acc.push_back(ProdSupport("Pedro", "Villasmil", 80000));

    for(size_t i=0; i<acc.size(); i++){
        cout << acc[i].getters() << endl;
        cout << acc[i].getAccess() << endl;
    }

    return 0;
}

C ++处理我自己的Exception类

2 个答案: