谁能用C语言给我写一个可以计算加减乘除的程序（只用算整数就OK）

Question

谁能用C语言给我写一个可以计算加减乘除的程序（只用算整数就OK）

Answer 1

int main()
{
int a,b,result;
char operator ;
printf("请输入两个数,输入的两个数用空格分隔\n");
scaf("%d %d",&a,&b);
printf("请指定你要输入的运算+或1表示加，-或2表示减，*或3表示乘，/或4表示除！\n") ;
scanf("%c",&operator) ;
switch operator
{
case ‘1’：‘+’
result=a+b;
break ;
case ‘2’：‘-’
result=a-b;
break ;
case ‘3’：‘*’
result=a*b;
break ;
case ‘4’：‘/’
result=a/b;
break ;
defaut:
break ;
}

printf("%d%c%d=%d\n",a,operator,b,result);

return 0 ;
}

Answer 2

#include <stdio.h>

int plus(int a, int b)
{
return a + b;
}

int minus(int a, int b)
{
return a - b;
}

int mutipe(int a, int b)
{
return a * b;
}

int div(int a, int b)
{
return a / b;
}

int main()
{
printf("下面为加减乘除法的测试:\n");
printf("%d, ",plus(2, 2));
printf("%d, ",minus(2, 2));
printf("%d, ",mutipe(2, 2));
printf("%d, ",div(2, 2));
return 0;
}

Answer 3

int main()
{
int a,b,result;
char operator ;
printf("请输入两个数,输入的两个数用空格分隔\n");
scaf("%d %d",&a,&b);
printf("请指定你要输入的运算+或1表示加，-或2表示减，*或3表示乘，/或4表示除！\n") ;
scanf("%c",&operator) ;
switch operator
{
case ‘1’：‘+’
result=a+b;
break ;
case ‘2’：‘-’
result=a-b;
break ;
case ‘3’：‘*’

Answer 4

我这程序的优点是当你还没输入=号之前,这程序是不会退出的,这样就可以实现你的多个数据的输入,用起来也很方便
#include<stdio.h>
main()
{
int a[20],i=0,result=0;
char ch;
printf("please input int num and char:\n");
while(ch!='=')
{
scanf("%d %c",&a[i],&ch);
switch(ch)
{
case '+': result=result + a[i]; break;
case '-': result=result - a[i]; break;
case '*': result=result * a[i]; break;
case '/': result=result / a[i]; break;
default:break;
}
i++;
}
printf("the result is:",result);
}

Answer 5

#include "stdio.h"

main()
{ int a,b,c; /*a,b是两个操作数，c是计算结果，ch是运算符*/
char ch; /*只有整形数据才可以求模运算，所以这里我将它们都设为整型*/
printf("Please input the expression a+b with no space between 'a' and '+' \n");
scanf("%d%c%d",&a,&ch,&b);/*Please input such as 3+5*/
switch(ch)
{ case '+': c=a+b; printf("\n%4d %c %4d=%4d\n",a,ch,b,c); break;
case '-': c=a-b; printf("\n%4d %c %4d=%4d\n",a,ch,b,c); break;
case '*': c=a*b; printf("\n%4d %c %4d=%4d\n",a,ch,b,c); break;
case '/': if(b==0) printf("error\n"); break;
else c=a/b; break;
case '%': if(b==0) printf("error\n"); break;
else c=a%b; break;
}

}

Answer 6

#include "stdio.h"

main()
{ int a,b,c; /*a,b是两个操作数，c是计算结果，ch是运算符*/
char ch; /*只有整形数据才可以求模运算，所以这里我将它们都设为整型*/
printf("Please input: \n");
scanf("%d%c%d",&a,&ch,&b);/*Please input */
switch(ch)
{ case '+': c=a+b; printf("\n%4d %c %4d=%4d\n",a,ch,b,c); break;
case '-': c=a-b; printf("\n%4d %c %4d=%4d\n",a,ch,b,c); break;
case '*': c=a*b; printf("\n%4d %c %4d=%4d\n",a,ch,b,c); break;
case '/': {if(b==0) {printf("error\n"); break; } else { c=a/b; printf("\n%d / %d=%d",a,b,c);} break;}
case '%': {if(b==0) {printf("error\n"); break; } else { c=a%b; printf("\n%d %% %d=%d",a,b,c);} break; }
}

}