c++从函数类型创建变量

C++ creating a variable from a function type

本文关键字：创建 变量 类型 函数 c++      更新时间：2023-10-16

我正试图将我的student_info结构与我的read_name功能联系起来，但我有问题，让它正常工作，它不会编译。我现在得到的错误是error: ‘first_name’ was not declared in this scope和error: ‘last_name’ was not declared in this scope。但是，我在结构中声明了它们。

下面是我的代码:

#include <iostream>
using namespace std;
//Place your structure here for Step #1:
struct student_info 
{
  char first_name[15];
  char last_name[15];
  char crn[15];
  char course_designator[15];
  int section;
};

//Place any prototypes that use the structure here:
void read_name(student_info & first_name[], student_info & last_name[])
{
  cout << "enter first name" << endl;
  cin.getline(first_name, 15, 'n'); 
  cout << "enter last name" << endl;
  cin.getline(last_name, 15, 'n');
  first[0] = toupper(first_name[0]);
  last[0] = toupper(last_name[0]); 
  cout << "your name is " << first_name << " " <<  last_name << endl;
}
int main()    
{
  //For Step #2, create a variable of the struct here:
  student_info student;
  read_name(first_name, last_name);
  return 0;
}

你可以做些什么来解决你的问题

  1. 将read_name更改为对student_info的引用

    void read_name(student_info & student)

  2. 修改read_name的实现，将数据读取到info的first_name和last_name成员。

    void read_name(student_info & student)
{
    cout << "enter first name" << endl;
    cin.getline(student.first_name, 15, 'n'); 
    cout << "enter last name" << endl;
    cin.getline(student.last_name, 15, 'n');
    first[0] = toupper(student.first_name[0]);
    last[0] = toupper(student.last_name[0]); 
    cout << "your name is " << student.first_name << " "
         <<  student.last_name << endl;
}

  3. 从main调用read_name，使用student作为参数。

    int main()    
{
    //For Step #2, create a variable of the struct here:
    student_info student;
    read_name(student);
    return 0;
}
相关文章：