使用函数计算学生的最终成绩答案

【问题标题】：Computing the final grade of a student using a function使用函数计算学生的最终成绩
【发布时间】：2015-09-19 09:16:11
【问题描述】：

我的任务：

学生最终成绩的计算考虑以下公式：FG = 3 个作业的 10% + 出勤率的 %10 + 3 个小测验的 15% + 3 个项目的 15% + 长测验的 20% + 30%大考。显示最终成绩和相应的备注。

要求：

允许用户重复操作或终止程序

备注：

90 - 100    Very Good

80-89       Good

70-79       Fair

60-69       Poor

Below 60    Needs Improvement

我的计划：

int average(int,int,int)

int a1,a2,a3,p1,p2,p3,s1,s2,s3,t1,l1,m1,FG;

intmain()

{

printf(" enter 3 assignments grade:  ");  scanf("%d%d%d",&a1,&a2,&a3);

printf(" enter 3 projects grade:   ");  scanf("%d%d%d",&p1,&p2,&p3);

printf(" enter 3 short quiz:     "); scanf("%d%d%d",&s1,&s2,&s3);

printf(" enter attendance:   "); scanf("%d",&t1);

printf(" enter long quiz grade:  "); scanf("%d",&l1);

printf(" enter major exam:   "); scanf("%d",&m1);

FG= average(a1,a2,a3,p1,p2,p3,s1,s2,s3,t1,l1,m1)

if((FG>=90) && (FG<=100))

printf("remark is VERY GOOD",FG);

if((FG>=80) && (FG<=89))

printf(" remark is GOOD",FG);

if((FG>=70) && (FG<=79))

printf(" remark is FAIR",FG);

if((FG>=60) && (FG<=69))

printf(" remark is POOR",FG);

printf(" [E]xit");

choice=getche();

switch(choice)

{

case 'e': 

case 'E': return 0;

default: printf(" press [E] to exit or press any key to repeat ");

getch();

main();
}

getch();

return 0;

}

void average( float a1, float a2, float a3, float p1, float p2, float p3, float s1, float s2,float s3,float t1,float l1, float m1)

{

return (((a1 + a2 + a3)/3)*.10) + (((p1 + p2 + p3)/3)*.15) + (((s1 + s2 + s3)/3)*.15) + (t1*.10) + (l1*.20) + (m1*.30);

}

疑问和问题：

发生错误是statement missing ; in function main
发生的错误是type mismatch in redeclaration of average
我获取平均值和函数调用的公式是否正确？

【问题讨论】：

请正确格式化您的问题。还有int average(int,int,int) --> int average(int,int,int,int,int,int,int,int,int,int,int,int);, intmain() --> int main(), void average( float a1, float a2, float a3, float p1, float p2, float p3, float s1, float s2,float s3,float t1,float l1, float m1) --> int average( int a1, int a2, int a3, int p1, int p2, int p3, int s1, int s2,int s3,int t1,int l1, int m1)
非常感谢您的帮助！ :)
请仔细阅读警告信息和错误，给点时间，你会自己找到答案的:)
现在我只有一个错误"语句丢失；在函数 main.
@Spencer ...所以添加缺少的;?

标签： c

【解决方案1】：

这是我制作的修改后的程序，它有效:)

int average(int,int,int,int,int,int,int,int,int,int,int,int);

int a1,a2,a3,p1,p2,p3,s1,s2,s3,t1,l1,m1,FG;

char choice;

int main()

{

clrscr();

printf(" enter 3 assignments grade: ");

scanf("%d%d%d",&a1,&a2,&a3);

printf(" enter 3 projects grade: ");

scanf("%d%d%d",&p1,&p2,&p3);

printf(" enter 3 short quizzes grade: ");

scanf("%d%d%d",&s1,&s2,&s3);

printf(" enter attendance grade: ");

scanf("%d",&t1);

printf(" enter long quiz grade: ");

scanf("%d",&l1);

printf(" enter major exam grade: ");

scanf("%d",&m1);

FG= average(a1,a2,a3,p1,p2,p3,s1,s2,s3,t1,l1,m1);

if((FG>=90) && (FG<=100))

printf("your remark is VERY GOOD",FG);

if((FG>=80) && (FG<=89))

printf("your remark is GOOD",FG);

if((FG>=70) && (FG<=79))

printf("your remark is FAIR",FG);

if((FG>=60) && (FG<=69))

printf("your remark is POOR",FG);

if(FG<60)

printf("your remark is NEEDS IMPROVEMENT",FG);

printf("\ndo you want to [E]xit?");

choice=getche();

switch(choice)

  {

    case 'e':

    case 'E': return 0;

    default: printf("invalid choice, press any key to repeat");

    getch();

    main();

  }

getch();

return 0;

}

int average(int a1,int a2,int a3,int p1,int p2,int p3,int s1,int s2,int s3,int 
t1,int l1,int m1)

{

return ((((a1 + a2 + a3)/3)*.10) + (((p1 + p2 + p3)/3)*.15) + (((s1 + s2 + s3)/3)*.15) + (t1*.10) + (l1*.20) + (m1*.30));

}

顺便说一句，感谢您之前的回答！ :) 让我耳目一新，呵呵 :)

【讨论】：

返回 ((((a1 + a2 + a3)/3)*.10) + (((p1 + p2 + p3)/3)*.15) + (((s1 + s2 + s3)/3)*.15) + (t1*.10) + (l1*.20) + (m1*.30));