从函数类型创建变量的C++

Question

我想链接我的student_info结构与我的read_name函数，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。然而，我在结构中声明了它们。

这里是我的代码：

#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; 
} 

Answer 1

事情可以做，以解决您的问题。

1. 变化read_name采取一个student_info的引用。

   ​​

2. 变化read_name的实施，将数据读入的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，使用student作为参数调用read_name。

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