Question

初始化我的char变量时遇到一些问题。

#include "stdafx.h"
#include <iostream>
using namespace std;
  class Scout {
    public:
     Scout(char Name,char Surname) : name(Name), surname(Surname) {
       cout << "Constructor 1" << " name = " << name << " surname = " <<surname;
   };
  private:
    char name, surname;
};
  int main()
 {
     Scout Andrew("Duck","Jones");
     system("pause");
     return 0;
  }

错误C2664'Scout :: Scout（Scout＆amp;＆amp;）'：无法将参数1从'const char [5]'转换为'char'

错误（活动）没有构造函数“Scout :: Scout”的实例与参数列表匹配

有没有其他方法来初始化姓名和姓氏？

Answer 1

构造函数的参数类型为char，而您在此处传递的参数Andrew("Duck","Jones");的类型为const char*。这不起作用。因为这是C ++，所以您可以使用std::string：

#include "stdafx.h"
#include <iostream>

  class Scout {
    public:
     Scout(std::string Name, std::string Surname) : name(Name), surname(Surname) {
       std::cout << "Constructor 1" << " name = " << name << " surname = " <<surname;
   };
  private:
    std::string name, surname;
};

int main()
{
   Scout Andrew("Duck","Jones");
   system("pause");
   return 0;
}

char初始化的问题

1 个答案: