使用栈实现整型表达式的计算

(C语言算法)顺序栈实现整型算术表达式的求值

//author：邵俊平
#include <stdio.h>
#include <stack>
#include <iostream>
using namespace std;




stack<int> num;
stack<char> operate;
int opp(int x1, int x2, char c)
{
switch(c)
{case '+':
return x1 + x2;

case '-':
return x1 - x2;
case '*':
return x1 * x2;
case '/':
return x1 / x2;


}


}
int calculate(char *s)
{
    char *p = s;
    while(*p)
   {
      if(*p>='0' && *p<='9')
           num.push(*p++ - '0');
      else if(*p == ')')
          {
 int x1 = num.top();
 num.pop();
 int x2 = num.top();
 num.pop();
 char op = operate.top();
 operate.pop();
 operate.pop();
 x1 = opp(x2, x1, op);
 num.push(x1);
 p++;
         }
 else if(*p == '*' || *p == '/')
 {
 char c = *p++;
 int x2 = *p++ - '0';
// p++;
 int x1 = num.top();
 num.pop();
 x1 = opp(x1, x2, c);
 num.push(x1);
 }
 else if(*p == '+' || *p == '-' || *p == '(')
 {
 operate.push(*p++);
 }
}
while(!operate.empty())
{
char c = operate.top();
operate.pop();
int x1 = num.top();
num.pop();
int x2 = num.top();
num.pop();
x1 = opp(x1, x2, c);
num.push(x1);
}
int x = num.top();
num.pop();
return x;
}


int main()
{
/*cout << sizeof(short) << sizeof(int) << endl;


int x = opp(4, 2, '-');
cout << x << endl;*/


int x = calculate("4*5+6/3");
cout << x << endl;
return 0;
}

/* 下面是别人的，看着好啰嗦啊~~~~~~

此程序的功能是求出用户输入的整形表达式的值20-10-07 17:43*/
/*输入的表达式请用#结尾，比如：2+3+5，应输入：2+3+5#。*/
#include <stdio.h>
#define   MAXSIZE 16

typedef struct{
      int data[MAXSIZE];
      int top;
      int base;
    }seqstack;   /* 顺序栈的定义*/
/*以下为函数声明*/
void InitStack(seqstack *);        
int Empty(seqstack *);
void Push(seqstack *, int );
int Pop(seqstack *);
int GetTop(seqstack *); 
int Operate(int ,char ,int );
char Proceed(char ,char );
int In(char );
int EvalExpres(void); 
/* 定义两个栈分别存放运算符和操作数*/
seqstack StackR,StackD; 
/*主函数*/
int main()
{
int v; 
char ch;
while(1)
{
    printf("\t@@@@@@@@本程序的功能为：用顺序栈实现整型算术表达式的求值@@@@@@@@\n");
    printf("\t@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@-20-10-07 17:47-@@@@@@\n"); 
    v = EvalExpres();         
    printf("The result is:%d",v);
                 /*以下为程序控制*/         
   printf("\nInput 'q' to quit and ENTER run again:"); 
    do{   
         scanf("%c",&ch);      
      if(ch == 'q' || ch == 'Q')
     exit(0); 
     }while(ch!='\n');
system("cls"); 
}
return 0; 
}

void InitStack(seqstack *s)
{     s->top = 0;
      s->base = 0; 
}   /* 初始化栈*/

int Empty(seqstack *s)
{   if(s->top == s->base)
             return 1;
     else 
        return 0;
}    /* 判断栈是否为空*/

void Push(seqstack *s, int x)
{    
    if(s->top == MAXSIZE)
         {    printf("OVER FLOW!\n");
              exit(0);
         }
    else 
         {    s->data[s->top] = x;
              s->top++;
         }     
}    /* 进栈 */

int Pop(seqstack *s)
{    int e;
     if(Empty(s))
        {   printf("Under flow!\n");
             return 0;
        }   /* 下溢*/
     else 
        {   s->top--;
          e = s->data[s->top];
            return e;
        }
} /* 出栈*/

int GetTop(seqstack *s)    /*取栈顶元素*/
{   
    if(Empty(s))
        {   printf("Under flow!\n");
             return 0;
        } 
   else 
      return s->data[s->top-1];      
}

int EvalExpres(void)    /* 表达式求解函数*/
{
     int a,b,i=0,s=0;
     char c[80],r;
     InitStack(&StackR);
     Push(&StackR,'#');
     InitStack(&StackD);
     printf(" 请输入表达式并以‘#’结束：");
     gets(c);
     while(c[i]!='#' || GetTop(&StackR)!='#')
      {      
         if(!In(c[i]))   /* 判断读入的字符不是运算符 是则进栈*/
           { if(c[i] >= '0' && c[i] <= '9')
                { 
                  s += c[i]-'0';      
                  while(!In(c[++i]))    /*此处实现的功能为当输入的数字为多位数时*/
                  { s*=10;
                    s += c[i]-'0';
                  }
                Push(&StackD,s+'0'); 
                  s = 0;
                }
             else 
              { 
                printf("你输入的表达式有误!\n");
                  return 0; 
              }            
           } 
      else         
         switch(Proceed(GetTop(&StackR),c[i])) /* 此函数用来比较读取的运算符和栈顶运算符的优先级*/
           {
               case '<':   /* 栈顶的元素优先级高*/
                 Push(&StackR,c[i]);
                    i++;
                  break;
               case '=':   /* 遇到匹配的小括号时则去掉他*/
                 Pop(&StackR);
                 i++;
                  break;
               case '>': /* 栈顶的优先级低则出栈并将结果写入栈内*/
                 r = Pop(&StackR);
                 a = Pop(&StackD)-'0';
                 b = Pop(&StackD)-'0';
                 Push(&StackD,Operate(a,r,b)) ;
                  break;                
           }
      } 
return (GetTop(&StackD)-'0');   /* 返回运算结果*/ 
}

int In(char c)         /*问题2：解决In函数问题:判断C是否为运算符是返回1否则返回0*/
{
   char ch[7]={'+','-','*','/','#','(',')'};
   int i;
   for(i = 0; i < 7; i++)
          if(c == ch[i])
             return 1;
             
             return 0;            
}

char Proceed(char op,char c) /*op为栈顶元素，c为当前读入的运算符,比较二者的优先级*/
{                                                    /*问题1：解决Proceed函数*/
      char ch;              
      if(op=='(' && c==')' || op=='#' && c=='#' ) 
          ch = '=';
else
      if(op=='+' || op=='-')     /*栈顶元素为‘+’或‘-’的时候*/
        switch(c)
           {
              case '+':
              case '-':
              case ')':
              case '#': ch = '>'; break;
              case '*':
              case '/':
              case '(': ch = '<';
           }
else 
      if(op=='*' || op=='/')    /*栈顶元素为‘*’或‘/’的时候*/
        switch(c)
           {
             case '+':
             case '-':
             case '*':
             case '/':     
             case ')':
             case '#':    ch = '>'; break;
             case '(':    ch = '<'; 
           }
else
      if(op=='(')             /*栈顶元素为‘(’的时候*/
        switch(c)
           { 
             case '+':
             case '-':
             case '*':
             case '/': 
             case '(': ch = '<'; break; 
             case '#':   printf("Error!\n"); exit(0);
           }
else
      if(op==')')     /*栈顶元素为‘)’的时候*/
        switch(c)
           {
             case '+':
             case '-':
             case '*':
             case '/': 
             case '#': ch = '>'; break; 
             case '(': printf("Error!\n"); exit(0);
           }
else 
      if(op=='#')              /*栈顶元素为‘#’的时候*/
        switch(c)
           {
             case '+':
             case '-':
             case '*':
             case '/': 
             case '(': ch = '<'; break; 
             case ')':   printf("Error!\n"); exit(0);
           } 
return ch;                               
}           
/*   问题3：解决Operate函数*/ 
int Operate(int a,char r,int b) /*返回由aRb的值 */
{ 
   int s;
   int d1 = a;
   int d2 = b; /*把字符ab变为对应数字*/
   switch(r)
      {
         case '+': s = d1+d2; break;
         case '-': s = d2-d1; break;
         case '*': s = d1*d2; break;
         case '/': s = d2/d1;    
      }
return (s+'0');    
}