Question

c语言编写 编写一个简单的计算器，实现两个整型数的四则运算。

c语言编写 编写一个简单的计算器，实现两个整型数的四则运算。

Answer 1

#include<iostream.h>
#include<stdio.h>
int main()
{ int a,b;
char ch,pm='Y';
while(pm!='N')
{
cout<<"请输入第一个数:"<<endl;
cin>>a;
cout<<"请输入运算符号:"<<endl;
cin>>ch;
cout<<"请输入第二个数:"<<endl;
cin>>b;
if(ch=='+')
cout<<a+b<<endl;
else if(ch=='-')
cout<<a-b<<endl;
else if(ch=='*')
cout<<a*b<<endl;
else if(ch=='/')
cout<<a/b<<endl;
cout<<"结束输入N，继续输入任意键！"<<endl;
pm=getchar();
}
return 1;}
要c++的行吗？还要换成c吗？

Answer 2

#include<stdio.h>
#include<stdlib.h>int main()
{
double num1,num2,result;
char op;
printf("请输入完整的数学运算符：\n");
scanf("%lf%c%lf",&num1,&op,&num2);
switch(op)
{
case '+':result = num1+num2;break;
case '-':result = num1 - num2;break;
case '*':result = num1 * num2;break;
case '/':result = num1/num2;break;
default:printf("运算符错误！");
}
printf("\n结果：%lf",result);
system("pause");}

Answer 3

#include<stdio.h>
int main()
{
int a,b;//存储两个整数
char c;
while(scanf("%d%c%d=",&a,&c,&b)!=EOF)
{
switch(c)
{
case'+':printf("%d\n",a+b);break;
case'-':printf("%d\n",a-b);break;
case'*':printf("%d\n",a*b);break;
case'/':printf("%d\n",a/b);break;
}
}
return 0;
}应该就是你所需要的吧

Answer 4

#include<stdio.h>void add(int a,int b);void minus(int a, int b);void multiply(int a,int b);void division(int a,int b);void main(){ add(6,4); minus(6,4); multiply(6,4); division(6,4);}void add(int a,int b){printf("%d+%d=%d\n",a,b,a+b);}void minus(int a,int b){printf("%d-%d=%d\n",a,b,a-b);}void multiply(int a,int b){printf("%d*%d=%d\n",a,b,a*b);}void division(int a,int b){//注意这里是整除，比如3/2=1，会舍掉小数点后面的所有数据printf("%d/%d=%d\n",a,b,a/b);}

Answer 5

这个……你打算如何实现呢？是读入算式？还是读入3个整数？