c语言表达式求值 只能算两个数的操作 超过了就有错误 麻烦大神看看怎么回事部分显示是我为了方便测试加的
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include<malloc.h>
#define size 100
#define size_1 10
typedef char Elemtype;
void InitStack ();
void Push();void function();
void Pop();
char sign(char ch1,char ch2);
char firstvalue(char ch);
typedef struct {
Elemtype *top;
Elemtype *base;
Elemtype stackSize;
}sqStack;
void InitStack (sqStack *s){
s->base=(Elemtype *)malloc(size * sizeof(Elemtype));
if(!s->base){
exit(0);
}
s->top=s->base;
s->stackSize=size;
}
void Push(sqStack *s,Elemtype e){
if(s->top - s->base >= s->stackSize){
s->base=(Elemtype *)realloc(s->base,(s->stackSize+size_1)* sizeof(Elemtype));
if(!s->base) exit(0);
}
*(s->top)=e;
s->top++;
}
void Pop(sqStack *s,char *e){
if(s->top == s->base)
return;
*e=*--(s->top);
}
int count(char b,char h,char a)
{
int i=0;
switch(a)
{
case '+':i=(h-'0')+(b-'0'); printf("执行了加法\n");;break;
case '-':i=(b-'0')-(h-'0'); printf("执行了减法\n");break;
case '*':i=(b-'0')*(h-'0'); printf("执行了乘法\n");break;
case '/':i=(b-'0')/(h-'0'); printf("执行了除法\n");break;
default:printf("错误\n");break;}
return i;
}
char firstvalue(char ch)
{
int k=0;
switch(ch)
{
case '+':k=0;break;
case '-':k=1;break;
case '*':k=2;break;
case '/':k=3;break;
case '(':k=4;break;
case ')':k=5;break;
case '#':k=6;break;
}
return k;}
char sign(char ch1,char ch2)
{
int i,j;
char a[7][7]={
{'>','>','<','<','<','>','>'},
{'>','>','<','<','<','>','>'},
{'>','>','>','>','<','>','>'},
{'>','>','>','>','<','>','>'},
{'<','<','<','<','<','=',' '},
{'>','>','>','>',' ','>','>'},
{'<','<','<','<','<',' ','='}
};
i=firstvalue(ch1);
j=firstvalue(ch2);//调用函数两次,返回两个函数值
return a[i][j];
}
int main(){
char a,e;
char n,c,b,k,h;
sqStack s1,s2;
InitStack(&s1);
InitStack(&s2);
Push(&s1,'#');
printf("请输入表达式,以#结尾\n");
c=getchar();
while(c!='#'||*(s1.top-1)!='#') {
if(c<='9'&&c>='0'){
Push(&s2,c);
c=getchar();
printf("已经压栈数字\n\n");
}
else
switch(sign(*(s1.top-1),c))
{
case'<':printf("已经执行1\n\n");Push(&s1,c);c=getchar();break;
case'=':Pop(&s1,&e);printf("已经执行9\n\n");break;
case'>':Pop(&s1,&a);Pop(&s2,&b);printf("b为:%c\n",b);Pop(&s2,&h);printf("h为:%c\n",h);n=count(b,h,a);printf("n为：%d \n\n",n);Push(&s2,n);printf("已经执行2 \n\n");break;
}}
Pop(&s2,&k);
printf("结果为:%d \n",k);
return 0;
}
// 3+5-9+7=6
// 3*4+2*6=24
// 3*(4+2)*6=108
// 3+5*9-7=41
// (3+5)*(9-7)=16
// 2*6*8/4=24
// 3*(4+8-7)*6=90
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include<malloc.h>
#define size 100
#define size_1 10
typedef char Elemtype;
//原代码void InitStack ();
//原代码void Push();void function();
//原代码void Pop();
char sign(char ch1,char ch2);
char firstvalue(char ch);
typedef struct
{
Elemtype *top;
Elemtype *base;
Elemtype stackSize;
}sqStack;
void InitStack (sqStack *s)
{
s->base=(Elemtype *)malloc(size * sizeof(Elemtype));
if(!s->base)
{
exit(0);
}
s->top=s->base;
s->stackSize=size;
}
void Push(sqStack *s,Elemtype e)
{
if(s->top - s->base >= s->stackSize)
{
s->base=(Elemtype *)realloc(s->base,(s->stackSize+size_1)* sizeof(Elemtype));
if(!s->base) exit(0);
}
*(s->top)=e;
s->top++;
}
void Pop(sqStack *s,char *e)
{
if(s->top == s->base)
return;
*e=*--(s->top);
}
//b是第1操作数,h是第2操作数
//例如,b - h = 7 - 1 = 6
// b / h = 8 / 4 = 2
int count(char b,char h,char a)
{
int i=0;
switch(a)
{
case '+':
i = b + h;
break;
case '-':
i = b - h;
break;
case '*':
i = b * h;
break;
case '/':
if(h==0) //除数不能为0
{
i=0;
}
else
{
i = b / h;
}
break;
default:
printf("错误\n");
break;
}
return i;
}
char firstvalue(char ch) //获得该运算符的优先级
{
int k=0;
switch(ch)
{
case '+':
k=0;
break;
case '-':
k=1;
break;
case '*':
k=2;
break;
case '/':
k=3;
break;
case '(':
k=4;
break;
case ')':
k=5;
break;
case '#':
k=6;
break;
default: //补充这个语句
k=7;
break;
}
return k;
}
char sign(char ch1,char ch2) //比较两个运算符的优先级
{
int i,j;
char a[7][7]={
{'>','>','<','<','<','>','>'},
{'>','>','<','<','<','>','>'},
{'>','>','>','>','<','>','>'},
{'>','>','>','>','<','>','>'},
{'<','<','<','<','<','=','&'},
{'>','>','>','>','&','>','>'},
{'<','<','<','<','<','&','='}};
//将数组a[][]里的空格' '改为'&"
//原代码
/*
char a[7][7]={
{'>','>','<','<','<','>','>'},
{'>','>','<','<','<','>','>'},
{'>','>','>','>','<','>','>'},
{'>','>','>','>','<','>','>'},
{'<','<','<','<','<','=',' '},
{'>','>','>','>',' ','>','>'},
{'<','<','<','<','<',' ','='}};
*/
i=firstvalue(ch1); //栈顶[*(s1.top-1)]"运算符"的优先级
j=firstvalue(ch2); //当前[c]"运算符"的优先级
if(i>6 || j>6) //补充这个语句
{
return '&'; //输入的"运算符"有错误
}
return a[i][j];
}
int main()
{
char a,e;
char n,c,b,k,h;
sqStack s1,s2; //s1运算符栈 s2数字栈
InitStack(&s1);
InitStack(&s2);
Push(&s1,'#');
printf("请输入表达式,以#结尾\n");
c=getchar();
while(c!='#'||*(s1.top-1)!='#')
{
if(c<='9'&&c>='0')
{
//将ASCII转换为数字
//这一步,一定要提前完成,不要让函数count来完成
c=c-'0';
Push(&s2,c);
c=getchar();
}
else
{
char preOp;
char whatSign;
preOp=*(s1.top-1); //[运算符]栈顶的运算符
whatSign=sign(preOp,c); //两个运算符的等级对比
switch(whatSign)
{
case'<':
Push(&s1,c);
c=getchar();
break;
case'=':
Pop(&s1,&e);
c=getchar(); //补充这个语句
break;
case'>':
Pop(&s1,&a); //运算符出栈:运算符=a
//原代码Pop(&s2,&b);
Pop(&s2,&h); //数字出栈:第2操作数=h
//原代码Pop(&s2,&h);
Pop(&s2,&b); //数字出栈:第1操作数=b
n=count(b,h,a);
Push(&s2,n); //中间的计算结果,重新入数字栈
break;
case '&': //补充这个语句
printf("\n输入的表达式有错误.\n");
return 0;
} //switch结束
} //if(c<='9'&&c>='0')else结果
} //while循环结束
Pop(&s2,&k);
printf("结果为:%d \n",k);
return 0;
}
