无法在 return 中将字符串转换为 int。在 int returning 函数 getsocial() 中没有声明任何字符串变量

Question

#include<iostream>
#include<string>
using namespace std;
/*this program takes a bank customer's info with the use of a class*/
//class definition
class BankCustomer
{
public:
    BankCustomer();     //constructor for BankCustomer class
    void fullname(string, string);
    string firstname();
    string lastname();
    bool setsocial(int s); //accept a arg. of int type
    int getsocial();
private:
    string fname, lname, SSNlength; //can't be changed by client; sensitive info should be made private
};

//class implementation
BankCustomer::BankCustomer(){}

void BankCustomer::fullname(string f, string l)
{
fname=f;
lname=l;
}

string BankCustomer::firstname()
{
return fname;
}

string BankCustomer::lastname()
{
return lname;
}

bool BankCustomer::setsocial(int s)//function verifies that entered SSN is 9 digits long by counting # of digits of entered SSN
{
int count, SSNlength;
while(s != 0)
{
s /=10;
++count;
if(count == 9)
{
cout <<"\nValid SSN Entered!" << endl;
SSNlength=s;
return true;
}
}
}

int BankCustomer::getsocial()
{
return SSNlength;
}

//client program
int main()
{
BankCustomer customer;          //customer declared as object of BankCust class
string firstname, lastname;
int ssn, s;

//data assignment
cout <<"\n Enter First Name\n" << endl;
cin >> firstname;

cout<<"\n Enter Last Name\n"<< endl;
cin >> lastname;

customer.fullname(firstname,lastname);

do
{
cout<<"\nEnter 9-Digit SSN"<< endl;
cin >> ssn;
customer.setsocial(ssn);
}
while(!customer.setsocial(ssn)); //function will repeat as long as entered user ssn forces setsocial() to evaluate it as false

//data ouput
cout <<"\nFirst Name:  "<<customer.firstname()<<"\n"<< endl;
cout <<"\nLast Name:  "<<customer.lastname()<<"\n"<< endl;
cout <<"\n SSN is:  "<<customer.getsocial()<<"\n" << endl;
}

错误针对 BankCustomer::getsocial()。变量 SSNlength 声明为 int 类型，getsocial() 具有 return 类型的 int。我在我的代码中没有看到我打算将 SSNlength 转换为字符串的任何地方。这就是我使用 setsocial() 函数传递和处理数据的方式吗？

Answer 1

错了：

int BankCustomer::getsocial()
{
return SSNlength;
}

因为 SSNlength 是类型 string class BankCustomer 的成员变量，从 BankCustomer::getsocial 可见，即使未在方法中声明。这就是 C++ 的工作原理。

因此编译器抱怨它无法将字符串转换为 int（您想要的 return）。

这是你声明它的地方：

private:
    string fname, lname, SSNlength; //can't be changed by client; sensitive info should be made private

鉴于变量的名称，我怀疑你想写：

private:
    string fname, lname;
    int SSNlength; //can't be changed by client; sensitive info should be made private

小心，在另一个方法中有一个自动变量以相同的方式调用。

我建议为成员制定一个命名规则来避免这一切。示例：为所有成员添加前缀 m_:

private:
    string m_fname, m_lname
    int m_SSNlength; //can't be changed by client; sensitive info should be made private

无法在 return 中将字符串转换为 int。在 int returning 函数 getsocial() 中没有声明任何字符串变量

Can't convert string to int in return. No string variable declared whatsoever in int returning function getsocial()

c++

class

function

data-conversion