写一个简单的程序 | 写一个简单的程序

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问题描述

编写程序以输入名称和卷号以及两个主题的标记，然后找到总数和百分比，然后打印名称，卷号，总数和百分比。

我尝试了什么：

Write a Program to enter name and roll number and marks of two subject and then find the total and percentage then print the name , roll number, total and percent.

What I have tried:

/* Program Of School */
#include <stdio.h>
void main ()
{
 char Name[50];
 Int Roll, Mark1, Mark2;
 float per;
Printf("enter the name and roll number\n");
scanf("%c %d" , &name, &Roll);
printf("enter the marks of two subject");
scanf("%d %d" , &mark1, &mark2);
sum=mark1+mark2:
per=(mark1+mark2)/2.0;
printf("the name is %c and roll mo is %d, &name, &roll number);
printf("total marks are %d and percentage are %f, &sum, &per);
getch();
}

推荐答案

带 scanf（）您必须将指针作为参数传递以获取存储的值。但是使用 printf（），你必须传递整数值，如 int 和 float ）按值。
％c格式序列适用于单个字符。使用％s获取字符串。
使用字符串（ char arrays）对象已经是一个指针。所以你必须传递普通名称或第一个元素的地址（名称或& Name [0] ）。
您的百分比计算错误。通用公式是 100 * part / total 。

With scanf() you have to pass pointers as arguments to get the values stored. But with printf() you have to pass integral values like int and float) by value.
The "%c" format sequence is for a single character. Use "%s" for strings.
With strings (char arrays) the object is already a pointer. So you have to pass the plain name or the address of the first element (Name or &Name[0]).
Your percentage calculation is wrong. The general formula is 100 * part / total.

#include <stdio.h>
int main ()
{
  char name[50];
  int roll, mark1, mark2;
  int sum;
  double avg;
  printf("enter the name and roll number\n");
  scanf("%s %d" , name, &roll);
  printf("enter the marks of two subject\n");
  scanf("%d %d" , &mark1, &mark2);
  sum = mark1 + mark2;
  avg = (mark1 + mark2)/2.0;
  printf("the name is %s and roll mo is %d\n", name, roll );
  printf("total marks are %d and the average is %g\n", sum, avg);
  getchar();
  return 0;
}

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