Linux 练习二

实现一个类似于计算器的程序

#include <stdlib.h>
#include <stdio.h>
#include <math.h>
int caltest(int a,char fuhao,int b);
int main(int argc, char*argv[])
{
int i = 0;
int a = 0;
int b = 0;
int result = 0;
if(argc < 4)
{
perror("please input over 4\n");
exit(1);
}

for(i = 0; i< argc; i++)
{
printf("argv[%d]=%s\n",i,argv[i]);
}
a = atoi(argv[1]);
b = atoi(argv[3]);
result = caltest(a,*argv[2],b);
printf("%s %s %s = %d\n",argv[1],argv[2],argv[3],result);
}

int caltest(int a,char fuhao,int b)
{
int result = 0;
switch(fuhao)
{
case '+':
result = a + b;
break;
case '-':
result = a - b;
break;
case '*':
result = a * b;
break;
case '/':
result = a / b;
break;
case '^':
result = (int)pow((double)a,(double)b);
break;
default:
result = 0;
break;
}
return result;
}

注意：在Linux中，"*"是一个通配符，代表所有字符，所以"*"必须要使用转义字符"\"