Question

我是该语言的新手，所有这些溢出问题和整数类型让我感到紧张。这就是我所拥有的，但是当我运行它时，我得到了，

-bash: syntax error near unexpected token `newline'

代码：

#include <stdio.h>
int main(void)
{
   int one, two, s, q, m;
   s = one+two
   q = one/two
   m = one*two
   printf("Enter first positive integer: ");
   scanf("%d", &one);
   printf("Enter second positive integer: ");
   scanf("%d", &two);
   printf("The addition of %d and %d is %d", one, two, s);
   printf("The integer division of %d divided by %d is %d", one, two, q);
   printf("the multiplication of %d and %d is %d", &one, &two, m);
   return 0;
}

谢谢

Answer 1

您应该在获得输入后执行计算。

printf("Enter first positive integer: ");
scanf("%d", &one);
printf("Enter second positive integer: ");
scanf("%d", &two);

s = one+two;
q = one/two;
m = one*two;

Answer 2

试试这个：

#include <stdio.h>
int main(void)
{
int one, two;
printf("Enter first positive integer: ");
scanf("%d", &one);
printf("Enter second positive integer: ");
scanf("%d", &two);
printf("The addition of %d and %d is %d", one, two, (one+two));
printf("The integer division of %d divided by %d is %d", one, two, (one/two));
printf("the multiplication of %d and %d is %d", &one, &two, (one*two));
return 0;
}

Answer 3

之后你缺少分号

   s = one+two
   q = one/two
   m = one*two

另外，您应该在阅读输入后执行计算，但这是一个不同的问题。

Answer 4

这些行会产生问题：

s = one+two
q = one/two
m = one*two

您错过了;（分号）

将其更改为：

s = one+two;
q = one/two;
m = one*two;

在执行操作之前，还要先阅读用户的输入。

编写一个程序，在C中打印两个输入整数的和，乘积和商

4 个答案: