逆波兰表达式的C实现

复习下数据结构，用栈简单实现逆波兰表达式，参考文档：

http://www.nowamagic.net/librarys/veda/detail/2307
http://www.nowamagic.net/librarys/veda/detail/2306

直接上代码：

/**
*code by lichmama from cnblogs.com
*@逆波兰表达式的C实现
*算术支持的运算模式：
*   四则运算；
*   操作数不得大于9；
*   中间数（即运算当中的临时变量）不得超过127
**/
#include <stdio.h>
#include <stdlib.h>

typedef struct __STACK__ {
    char op;
    struct __STACK__ *next;
}STACK, *PSTACK;

void init(PSTACK *);
void push(PSTACK *, char);
void pop(PSTACK *, char *);
void clear(PSTACK *);
void destroy(PSTACK *);

char getoplevel(char op){
    if(op=='+' || op=='-')
        return ;
    if(op=='*' || op=='/')
        return ;
    return ;
}

char calc(char *rpn){
    char *p=rpn;
    char e;
    int x, y;
    PSTACK pStack;
    init(&pStack);
    while(*p){
        if(<=*p && *p<=){
            push(&pStack, *p);
        }else {
            pop(&pStack, &e);x=e;
            pop(&pStack, &e);y=e;
            switch(*p){
                case '+':push(&pStack, y+x);break;
                case '-':push(&pStack, y-x);break;
                case '*':push(&pStack, y*x);break;
                case '/':push(&pStack, y/x);break;
            }
        }
        p++;
    }
    pop(&pStack, &e);
    free(pStack);pStack=NULL;
    return e;
}

int main(void){
    char e;
    char old_exp[]="(3-1)*3+8/2+(9*3/(2+1)+3*4/6)-2=";
    char rpn_exp[]="";
    char *p=old_exp;
    char *r=rpn_exp;
    PSTACK gStack;
    init(&gStack);
//
    while(*p!='\0' && *p!='='){
        if(''<=*p && *p<=''){
            *r++=(*p-'');
        }else if(*p=='('){
            push(&gStack, '(');
        }else if(*p==')'){
            for(;;){
                pop(&gStack, &e);
                if(e=='(')break;
                *r++=e;
            }
        }else if(*p=='+' || *p=='-' || *p=='*' || *p=='/'){
            if(getoplevel(gStack->op)<getoplevel(*p)){
                push(&gStack, *p);
            }else{

                for(;;){
                    pop(&gStack, &e);
                    if(getoplevel(e)<getoplevel(*p)){
                        if(e!='#')push(&gStack, e);
                        push(&gStack, *p);break;
                    }
                    *r++=e;
                }
            }
        }
        p++;
    }
    for(;;){
        pop(&gStack, &e);
        if(e=='#')break;
        *r++=e;
    }
    //printf("%s\n", rpn_exp);
    printf("%d\n", calc(rpn_exp));
//
    clear(&gStack);
    destroy(&gStack);
    return ;
}

void init(PSTACK *s){
    *s=(PSTACK)malloc(sizeof(STACK));
    (*s)->op='#';
    (*s)->next=NULL;
}

void push(PSTACK *s, char e){
    PSTACK p=(PSTACK)malloc(sizeof(STACK));
    p->op=e;
    p->next=*s;
    *s=p;
}

void pop(PSTACK *s, char *e){
    if((*s)->next){
        PSTACK p=(*s);
        *e=(*s)->op;
        *s=(*s)->next;
        free(p);
        p=NULL;
    }else *e='#';
}

void clear(PSTACK *s){
    PSTACK p;
    while((*s)->next){
        p=*s;
        *s=(*s)->next;
        free(p);
        p=NULL;
    }
}

void destroy(PSTACK *s){
    free(*s);
    *s=NULL;
}

注：改善操作数的限制可使用整形（或浮点型）数组存放中间变量。