栈的应用之求逆波兰表达式的值

Posted 2020-06-17

  1 #include<stdio.h>
  2 #include<stdlib.h>
  3 #include<ctype.h>
  4 
  5 #define OK         1
  6 #define ERROR         0
  7 #define STACK_INIT_SIZE 20
  8 #define STACK_INCREMENT 10
  9 #define DIGITBUFFER     10
 10 
 11 typedef int    Status;
 12 typedef double Elemtype;
 13 typedef struct StackNode{
 14     Elemtype* base;
 15     Elemtype* top;
 16     int stackSize;
 17 }StackNode;
 18 typedef struct StackNode* Stack;
 19 
 20 Status InitStack(Stack s){
 21     s->base = (Elemtype*)malloc(sizeof(Elemtype) * STACK_INIT_SIZE);
 22     if(!s->base)
 23         return ERROR;
 24     s->top = s->base;
 25     s->stackSize = STACK_INIT_SIZE;
 26     return OK;
 27 }
 28 Status Pop(Stack s,Elemtype* result){
 29     if(s->base == s->top)
 30         return ERROR;
 31     *result = *(--s->top);
 32     return ERROR;
 33 }
 34 Status Push(Stack s,Elemtype value){
 35     if(s->top - s->base == s->stackSize){
 36         s->base = (Elemtype*)realloc(s->base,sizeof(Elemtype) * (STACK_INIT_SIZE + STACK_INCREMENT));
 37         if(!s->base)
 38             return ERROR;
 39         s->top = s->base + STACK_INIT_SIZE;
 40         s->stackSize = STACK_INIT_SIZE + STACK_INCREMENT;
 41     }
 42     *(s->top) = value;
 43     s->top++;
 44     return OK;
 45 }
 46 int StackLenth(Stack s){
 47     return s->top - s->base;
 48 }
 49 Status RPT(){            //reverse polish notation
 50     char c;
 51     double operater1,operater2;
 52     double result;
 53     int i = 0;
 54     char bufferDigit[DIGITBUFFER];
 55 
 56     Stack s;
 57     InitStack(s);
 58 
 59     printf("  Please Enter Reverse Polish Notation!(RPN)\n");
 60     printf("------note:  separated by space between -------\n");
 61     printf("------     number or operator.end of ‘#‘-------\n");
 62 
 63     scanf("%c", &c);
 64     while(c != ‘#‘){
 65         /* 处理输入的数字：由于使用%c接受输入，所以对于123这样的多位数的
 66          * 输入%c是不能处理的。所以设置了char型的数组bufferDigit来缓存输
 67          * 入的多位数。在开始了多位数的输入之后，必将以空格来结束该多位数
 68          * 输入，所以if(c == ‘ ‘)用来判断多位数的结束。以便将缓存在char
 69          * 型数组中的多位数转化为double并且Push。
 70          */
 71         while( isdigit(c) || c == ‘.‘){
 72             if(i == 10){
 73                 printf("number is too lager\n");
 74                 return ERROR;
 75             }
 76             bufferDigit[i++] = c;
 77             bufferDigit[i] = ‘\0‘;
 78             scanf("%c", &c);
 79             if(c == ‘ ‘){    //不是空格就一定还是数字
 80                 result = atof(bufferDigit);
 81                 Push(s,result);
 82                 i = 0;
 83             }
 84         }
 85         /* 处理输入的运算符
 86          */
 87         switch(c){
 88             case ‘+‘:
 89                 Pop(s,&operater1);
 90                 Pop(s,&operater2);
 91                 Push(s,operater1 + operater2);
 92                 break;
 93             case ‘-‘:
 94                 Pop(s,&operater1);
 95                 Pop(s,&operater2);
 96                 Push(s,operater2 - operater1);
 97                 break;
 98             case ‘*‘:
 99                 Pop(s,&operater1);
100                 Pop(s,&operater2);
101                 Push(s,operater1 * operater2);
102                 break;
103             case ‘/‘:
104                 Pop(s,&operater1);
105                 Pop(s,&operater2);
106                 Push(s,operater2 / operater1);
107                 break;
108         }
109         scanf("%c", &c);
110     }
111 
112     Pop(s,&result);
113     printf("The result of RPN is %f\n",result);
114 }
115 
116 //test;
117 Status ShowStack(Stack s){
118     while(s->base != s->top){
119         printf("%f ",*(--(s->top)));
120     }
121     printf("\n");
122 }
123 Status Test(){
124     Stack s1;
125     InitStack(s1);
126     Push(s1,1);
127     Push(s1,2);
128     Push(s1,3);
129     ShowStack(s1);
130 }
131 int main(){
132     RPT();
133 
134     Stack s;
135     InitStack(s);
136     Push(s,1);
137     Push(s,2);
138     Push(s,3);
139     ShowStack(s);
140     return 0;
141 }
142 /* 该程序编写的RPT部分功能可以完成，但是Text部分存在不解之处，在程序的124~129行的代码与134到
143  * 139行的代码完全相同。但是在main函数中的可以顺利运行，而在Test函数中的则出现错误(无错误提示)
144  * 如果将124~129行的代码改成如下则可以顺利运行：
145     StackNode s1;
146     InitStack(&s1);
147     Push(&s1,1);
148     Push(&s1,2);
149     Push(&s1,3);
150     ShowStack(&s1);
151    不知为何期待大牛指导。
152  * */