急！双栈法实现算术表达式的求值演示，以下代码如何扩展到实数

Question

#include<stdio.h>

#include<stdlib.h>

#define OK 1

#define ERROR 0

#define STACKINCREMENT 5

#define STACK_INIT_SIZE 10

typedef char SElemType;

typedef int Status;

typedef struct{

SElemType *base;//栈底指针

SElemType *top;//栈顶指针

int stacksize;//当前已经分配的存储空间

}SqStack;

char precede[7][7]={

    {'>','>','<','<','<','>','>'},

    {'>','>','<','<','<','>','>'},

    {'>','>','>','>','<','>','>'},

    {'>','>','>','>','<','>','>'},

    {'<','<','<','<','<','=','!'},

    {'>','>','>','>','!','>','>'},

    {'<','<','<','<','<','!','='}

};//定义算符之间优先关系的二维数组

//构造一个存放char型数据的空栈

Status InitStack(SqStack *s){

s->base = (SElemType *)malloc(STACK_INIT_SIZE*sizeof(SElemType));

if(!s->base) return ERROR;

s->top = s->base;//栈中元素个数为0

s->stacksize = STACK_INIT_SIZE;

return OK;

}

//入栈

Status Push(SqStack *s,SElemType e){

if(s->top-s->base>=s->stacksize){

s->base = (SElemType *)realloc(s->base,(STACKINCREMENT+s->stacksize)*sizeof(SElemType));

if(!s->base) exit(0);

s->top = s->base+s->stacksize;

s->stacksize += STACKINCREMENT;

}

*s->top++ = e;

return OK;

}

//出栈

Status Pop(SqStack *s,SElemType *e)

{

if(s->base==s->top){

printf("空栈！\n");

return ERROR;

}

*e = *--s->top;

return OK;

}

//得到栈顶元素

SElemType GetTop(SqStack *s){

return *(s->top-1);

}

//确定输入的字符如果是操作符的话判断在二维数组中的下标 若是数字的话就另外与操作符区分开 便于在输入表达式时是入哪个栈

int Index(char c){

switch(c){

case '+': return 0;

case '-': return 1;

case '*': return 2;

case '/': return 3;

case '(': return 4;

case ')': return 5;

case '#': return 6;

default:  return 7;

}

}

//判断优先级,返回大小 < > = !

char Priority(char a,char b){

int x,y;

x = Index(a); y = Index(b);

if(x!=7&&y!=7)

return precede[x][y];

else

return '!';

}

//简单表达式求值

double Operate(double a,char theta,double b){

switch(theta){

case '+': return a + b;

case '-': return a - b;

case '*': return a * b;

case '/': return a / b;

}

}

//判断是字符是否是数字

Status isdigit(char ch){

if(ch>='0'&&ch<='9') return OK;

return ERROR;

}

//算术表达式求值

Status PrintStack1(SqStack S,int &m)

//若栈S不空则从栈底到栈顶输出S的元素(为char型)并返回打印长度

{

  m=0;

 char *p;

 if(S.base==S.top)  return ERROR;

 for(p=S.base;p!=S.top;p++)

 { printf("'%c'",*p); m++;}

 return OK;

}//PrintStack1

Status PrintStack2(SqStack S,int &n)

//若栈S不空则从栈底到栈顶输出S的元素(为char型)并返回打印长度

{

  n=0;

 char *p;

 if(S.base==S.top)  return ERROR;

 for(p=S.base;p!=S.top;p++)

 { printf("'%d'",*p); n++;}

 return OK;

}//PrintStack1

int GetExpressionValue(){

int m,n,i=0;

SqStack OPTR,OPND;

SElemType result;//返回最后结果

InitStack(&OPTR);

InitStack(&OPND);

Push(&OPTR,'#');//将结束符置于操作符的底端

printf("请输入算术表达式(以#结尾,负数用括号括起来）:\n");

char c = getchar();

printf("步骤\tOPTR栈\t\tOPND栈\t\t当前字符   主要操作\n");

while(c!='#'||GetTop(&OPTR)!='#'){//当*c=='#'&&栈顶字符=='#'的时候

i++;

printf("  %d\t",i);//打印序号

PrintStack1(OPTR,m);putchar('\t');//打印OPTR栈

if(m<10) putchar('\t');//若打印长度不够8要再

PrintStack2(OPND,n);putchar('\t');//打印OPND栈

if(n<10) putchar('\t');

if(isdigit(c)){//如果是数字的话将其转化为数字 然后入操作数栈

int data[10];

int i,num;

i = num =0;//num是一个中间数 用于将字符串中的数字转化为整数然后入栈 i是用于将字符串中的字符存入data数组

while(isdigit(c)){

data[i] = c-'0';

i++;

c = getchar();

}

for(int j=0;j<i;j++){

num = num*10+data[j];

}

printf("   %d\t  ",num);//打印当前字符

printf("Push2(OPND,%d)\n",num);//打印该操作

Push(&OPND,num);

}else{//如果是字符的话将其入操作符栈

SElemType a,b,theta;//a b theta是用来返回操作数栈和操作符栈里的元素的

switch(Priority(GetTop(&OPTR),c)){//比较即将入栈的字符与栈顶 操作符的优先级关系

case '<':Push(&OPTR,c);

printf("   %c\t  ",c);//打印当前字符

printf("Push1(OPND,%c)\n",c);//打印该操作

c = getchar();

break;

case '>':Pop(&OPND,&b);

Pop(&OPND,&a);

Pop(&OPTR,&theta);

Push(&OPND,Operate(a,theta,b));

printf("   %c\t  ",c);//打印当前字符

printf("Operate(%d,'%c',%d)\n",a,theta,b);//打印该操作

break;//将结果入栈

case '=':Pop(&OPTR,&theta);

c = getchar();

break;//说明括号相遇 删除栈内括号即可

default:break;

}

}

}

Pop(&OPND,&result);

printf("结果是：%d",result);

}

int main(){

GetExpressionValue();

return 0;

}

Answer 1

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

Answer 2

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Answer 3

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