栈实现的带括号的计算器



带括号的表达式计算：通过栈的应用来实现，建立操作数栈和运算符栈，运算符有优先级。
规则：1.自左至右扫描表达式，凡是遇到操作数一律进操作数栈。
 2.当遇到运算符时如果它的优先级比运算符栈栈顶元素的优先级高就进栈。反之，取出栈顶运算符和操作数栈栈顶
 的连续两个操作数进行运算，并将结果存入操作数栈，然后继续比较该运算符与栈顶运算符的优先级。
 3.遇到左括号一律进运算符栈，右括号一律不进运算符栈。在遇到对应的右括号前，括号内运算按规则2进行，

 当遇到右括号时，将括号内运算结果进操作数栈，并取出左括号。

calculate.c

#include<stdio.h>#include<stdlib.h>#define MAX_SIZE 100enum usu{STACK_FULL = -10,STACK_OK,STACK_EMPTY,SUCCESS,PUSH_SUCCESS};int calculate(int num1,int num2,char c){switch(c){case '+':{return num1 + num2;}case '-':{return num1 - num2;}case '*':{return num1 * num2;}case '/':{return num1 / num2;}}}int priority(char c)//计算运算符的优先级{switch(c){case '+':case '-':{return 1;break;}case '*':case '/':{return 2;break;}default:{return 0;}}return 0;}void initialize_stack(int *stack,int *top)//初始化顺序栈{int i;for(i = 0;i < MAX_SIZE;i ++){stack[i] = 0;}*top = -1;}void initialize_stack2(char *stack,int *top){int i;for(i = 0;i < MAX_SIZE;i ++){stack[i] = 0;}*top = -1;}int is_stack_full(int *top)//检查栈是不是满{if(*top == MAX_SIZE - 1){return STACK_FULL;}return STACK_OK;}int is_stack_empty(int *top)//检查栈是不是空{if(*top == -1){return STACK_EMPTY;}return STACK_OK;}int push_stack(int *stack,int *num,int *top)//入栈{if(is_stack_full(top) == STACK_FULL){return STACK_FULL;}(*top) ++;stack[*top] = *num;return PUSH_SUCCESS;}int push_stack2(char *stack,char *c,int *top){if(is_stack_full(top) == STACK_FULL){return STACK_FULL;}(*top) ++;stack[*top] = *c;return PUSH_SUCCESS;}int pop_stack(int *stack,int *top)//出栈{int num;if(is_stack_empty(top) == STACK_EMPTY){return STACK_EMPTY;}num = stack[*top];(*top) --;return num;}char pop_stack2(char *stack,int *top){char c;if(is_stack_empty(top) == STACK_EMPTY){return STACK_EMPTY;}c = stack[*top];(*top) --;return c;}#if 0比较解析的运算符+-*/（）与栈顶运算符的优先级,按照规则进行各种出入栈运算#endifint compare_priority(char *str,char *stack2,int *top2,int *stack1,int *top1){int num,num1,total;char c;if(*str == '\0')//文件尾时，正常退出{return SUCCESS;}if(is_stack_full(top2) == STACK_FULL){return STACK_FULL;}else if(is_stack_empty(top2) == STACK_EMPTY||*str == '('){push_stack2(stack2,str,top2);}else if(*str == ')'){c = pop_stack2(stack2,top2);//从栈顶取出一运算符while(c != '('){num = pop_stack(stack1,top1);//取出一操作数num1 = pop_stack(stack1,top1);total = calculate(num1,num,c);//保存运算结果push_stack(stack1,&total,top1);//将运算结果入栈c = pop_stack2(stack2,top2);}//出while循环即'('运算符已取出}else//解析出的运算符为+-*/时执行以下操作{c = pop_stack2(stack2,top2);if(c == '(')//运算符栈顶元素为'('时，解析出的运算符要入栈{push_stack2(stack2,&c,top2);push_stack2(stack2,str,top2);return SUCCESS;}num = priority(c);num1 = priority(*str);if(num1 > num)//解析出的运算符优先级大于栈顶元素优先级时，解析出的运算符要入栈{push_stack2(stack2,&c,top2);push_stack2(stack2,str,top2);}else{//取出栈顶运算符和操作数栈栈顶的连续两个操作数进行运算，//并将结果存入操作数栈，然后继续比较该运算符与栈顶运算符的优先级。num = pop_stack(stack1,top1);num1 = pop_stack(stack1,top1);total = calculate(num1,num,c);push_stack(stack1,&total,top1);compare_priority(str,stack2,top2,stack1,top1);}}return SUCCESS;}int main(){int num,num1,total;char c;char *dest = (char *)malloc(sizeof(100)); char *temp = dest;//记录分配空间的首地址char *str = (char *)malloc(sizeof(100));//保存需要计算的表达式字符串。int stack1[MAX_SIZE];//操作数栈int top1;initialize_stack(stack1,&top1);//初始化栈char stack2[MAX_SIZE];//运算符栈int top2;initialize_stack2(stack2,&top2);printf("please input the formula:\n");scanf("%s",str);char *ptr = str;while(*str != '\0'){if(*str >= '0'&&*str <= '9')//解析连续的数字字符为整数，并入操作数栈{while(*str >= '0'&&*str <= '9'){*temp = *str;str ++;temp ++;}//while循环结束，str指向运算符或指向字符串末尾*temp = '\0';temp = dest;num = atoi(dest);//字符串转整型push_stack(stack1,&num,&top1);}//比较解析的运算符+-*/（）与栈顶运算符的优先级,按照规则进行各种出入栈运算compare_priority(str,stack2,&top2,stack1,&top1);if(*str == '\0'){break;}str ++;}/*依次取出操作数栈顶两元素与运算符栈顶一元素进行计算，结果保存在操作数栈，直至运算符栈为空*/while(top2 != -1){num = pop_stack(stack1,&top1);num1 = pop_stack(stack1,&top1);c = pop_stack2(stack2,&top2);total = calculate(num1,num,c);push_stack(stack1,&total,&top1);}printf("%s=%d\n",ptr,total);return 0;}

编译执行结果如下：

[root@localhost calculate]# gcc calculate2.c &&./a.out
please input the formula:
3+4-(2*5)+19
3+4-(2*5)+19=16

[root@localhost calculate]# gcc calculate2.c &&./a.out
please input the formula:
3-(4+(5*2))+21
3-(4+(5*2))+21=10

本文转自：http://blog.csdn.net/yinshitaoyuan/article/details/51638878