c++ - 按值将类型结构传递给函数，错误 :two or more data types in declaration of 'average' !! -

Question

大家好

我有一个小问题(或者它可能很大!)，在编译时只有一个错误，但众所周知，一个错误就是阻碍进展的全部。 基本上我是 C++ 的新手，我的任务是编写以下代码并按值将类型 stuct 参数传递给函数。但我收到以下错误消息: “平均值声明中的两种或多种数据类型”，因此我将不胜感激对我的一个错误的任何解决方案。 非常感谢...

enter code here
#include <iostream> 
#include <cstdio>
#include <math.h>
using namespace std;


           struct student{

              char name[40];
              int student_id;
              int student_grades[3];
              int average;
           };

int main ()
{
   extern int average(student);
   student programming;

   int j;

         cout<<"\nPlease Enter the student name for student number: ";

         cin>>programming.name;

         cout<<"\nPlease Enter student i.d for student number: ";

         cin>>programming.student_id;

         cout<<"\nPlease Enter student grades for student number: ";

         for(j=0;j<3;j++){

                    cout<<"\nEnter student grade no: "<<j+1<<"\n";
                    cin>>programming.student_grades[j];               
         }

         programming.average=average(programming);

         cout<<"\nNo.      Name      ID Number      Average\n";
         cout<<programming.name;
         cout<<"      "<<programming.student_id <<"      ";
         cout<<programming.average<<"      ";             

     system ("PAUSE");
     return 0;
}

struct student;
int void average(student programming){

    int sum=0;
    int ave=0;
    int j;

    for(j=0;j<3;j++){
           sum=sum+programming.student_grades [j];
    }
    ave=sum/3;
    return ave;
}
enter code here

Answer 1

int void average(student programming) 不是有效语法。返回类型只有一种，即int average(student programming)。