Question

我正在尝试将我的student_info结构与我的read_name函数相关联，但我遇到了使其正常工作并且无法编译的问题。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

您可以采取哪些措施来解决问题。

更改read_name以引用student_info。
```
void read_name(student_info & student)
```
更改read_name的实施，将数据读入first_name的{{1}}和last_name成员。
```
info
```
从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; }开始，使用main作为参数调用read_name。
```
student
```

C ++从函数类型创建变量

1 个答案: