C语言计算简单表达式

这是一个简单的C语言计算机表达式，大家有时间可以研究一下！

#include <stdlib.h>
#include <stdio.h>
#include <conio.h>
#include <string.h>
int eadd(int x,int y){return x y;}
int esub(int x,int y){return x-y;}
int emul(int x,int y){return x*y;}
int ediv(int x,int y){return x/y;}
int (*func[])(int,int)={eadd,esub,emul,ediv};

int num,curch;
char chtbl[]=/" -*/()=/";
char corch[]=/" -*/()=0123456789/";

int getach()
{
int i;
while(1)
{
curch=getchar();
if(curch==EOF)return -1;
for(i=0;corch[i]&&curch!=corch[i];i );
if((unsigned int)i<strlen(corch))break;
}
return curch;
}

int getid()
{
int i;
if(curch>=’0’&&curch<=’9’)
{
for(num=0;curch>=’0’&&curch<=’9’;getach())
num=num*10 curch-’0’;
return -1;
}
else
{
for(i=0;chtbl[i];i )
if(chtbl[i]==curch)break;
if(i<=5)getach();
return i;
}
}

int cal()
{
int x1,x2,x3,op1,op2,i;
i=getid();
if(i==4)x1=cal();else x1=num;
op1=getid();
if(op1>=5)return x1;
i=getid();
if(i==4)x2=cal();else x2=num;
op2=getid();
while(op2<5)
{
i=getid();
if(i==4)x3=cal();else x3=num;
if((op1/2==0)&&(op2/2==1))
x2=(*func[op2])(x2,x3);
else
{
x1=(*func[op1])(x1,x2);
x2=x3;
op1=op2;
}
op2=getid();
}
return (*func[op1])(x1,x2);
}
void main(void)
{
int value;
printf(/"please input an expression://n/");
getach();
while(curch!=’=’)
{
value=cal();
printf(/"the result is:%d//n/",value);
printf(/"please input an expression://n/");
getach();
}
}