Question

大家好，我为我的课程创建了一个对象，并尝试将其输出为BOb，c和1900.543，但该程序的名称和类型为空白，平衡为0。任何人都可以帮助我，并告诉我我的程序输出不符合预期的问题是什么。

#include <iostream>
#include <vector>
#include <string>

using namespace std;

class Record{

private:

  string newName;
  char newType;
  double newBalance;

public:
  Record(string, char, double);
  Record();
  string getName();
  char getType();
  double getBalance();

};

Record::Record(string name, char type, double balance)
{
   balance = newBalance;
   name = newName;
   type = newType; 
} 
string Record::getName()
{
    return newName;
}
char Record::getType()
{
   return newType;
}
double Record::getBalance()
{
   return newBalance;
} 

int main()
{

  string name = "Bob";
  char type = 'c';
  double balance = 1900.543;


  Record c1(name, type, balance);
  cout << c1.getName() << endl;
  cout << c1.getBalance() << endl;
  cout << c1.getType() << endl;


 return 0;
}

Answer 1

在发布到论坛之前，请尝试调试。

Record::Record(string name, char type, double balance)
{
   balance = newBalance;
   name = newName;
   type = newType; 
}

将此更改为

Record::Record(string name, char type, double balance)
    {
       newBalance = balance;
       newName = name;
       newType = type; 
    }

Answer 2

之所以发生此错误，是因为编译器无法就此警告您。
为了避免这种情况，您可以使用类似以下的构造函数初始化器列表：

Record::Record(string name, char type, double balance): newBalance(balance), newName(name), newType(type)
{

}

假设您已像在代码中那样反转了成员变量和参数的位置：

Record::Record(string name, char type, double balance): balance(newBalance), name(newName), type(newType)
{

}

编译器会指出此错误：

error: class 'Record' does not have any field named 'balance'
 Record::Record(string name, char type, double balance): balance(newBalance), name(newName), type(newType)
                                                         ^~~~~~~
error: class 'Record' does not have any field named 'name'
 Record::Record(string name, char type, double balance): balance(newBalance), name(newName), type(newType)
                                                                              ^~~~
error: class 'Record' does not have any field named 'type'
 Record::Record(string name, char type, double balance): balance(newBalance), name(newName), type(newType)

c ++对象输出错误信息

2 个答案: