栈的应用之求逆波兰表达式的值

 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  * */