Exercises 4-3

Tips:此为逆波兰计算器实现

如：

(1 – 2) * (4 + 5)
用逆波兰表示法为：
1 2 - 4 5 + *

#include <stdio.h>
#include <stdlib.h> /* for atof() */
#include <math.h>
#define MAXOP 100 /* max size of operand or operator */
#define NUMBER '0' /* signal that a number was found */
int getop(char []);
void push(double);
double pop(void);
/* reverse Polish calculator */
main()
{
    int type;
    double op2;
    char s[MAXOP];
    while ((type = getop(s)) != EOF)
    {
        switch (type)
        {
        case NUMBER:
            push(atof(s));
            break;
        case '+':
            push(pop() + pop());
            break;
        case '*':
            push(pop() * pop());
            break;
        case '-':
            op2 = pop();
            push(pop() - op2);
            break;
        case '/':
            op2 = pop();
            if (op2 != 0.0)
                push(pop() / op2);
            else
                printf("error: zero divisor\n");
            break;
        case '%':
            op2 = pop();
            if (op2 != 0.0)
                push(fmod(pop(),op2));
            else
                printf("error: zero divisor\n");
            break;
        case '\n':
            printf("\t%.8g\n", pop());
            break;
        default:
            printf("error: unknown command %s\n", s);
            break;
        }
    }
    return 0;
}
#define MAXVAL 100 /* maximum depth of val stack */
int sp = 0; /* next free stack position */
double val[MAXVAL]; /* value stack */
/* push: push f onto value stack */
void push(double f)
{
    if (sp < MAXVAL)
        val[sp++] = f;
    else
        printf("error: stack full, can't push %g\n", f);
}
/* pop: pop and return top value from stack */
double pop(void)
{
    if (sp > 0)
        return val[--sp];
    else
    {
        printf("error: stack empty\n");
        return 0.0;
    }
}
#include <ctype.h>
int getch(void);
void ungetch(int);
/* getop: get next character or numeric operand */
int getop(char s[])
{
    int i, c;
    while ((s[0] = c = getch()) == ' ' || c == '\t')/*ignore white space*/
        ;
    s[1] = '\0';
    if (!isdigit(c) && c != '.'&&c!='-')
        return c; /* not a number */
    i = 0;
    if(c=='-')
    {
        if(isdigit(c=getch())||c=='.')
            s[++i]=c;
        else
        {
            if(c!=EOF)
                ungetch(c);
            return '-';
        }
    }
    if (isdigit(c)) /* collect integer part */
        while (isdigit(s[++i] = c = getch()))
            ;
    if (c == '.') /* collect fraction part */
        while (isdigit(s[++i] = c = getch()))
            ;
    s[i] = '\0'; //这一步可以将可以忽略掉上面操作数中的非数字，因为它并不属于操作数的一部分，比如123+,+是操作符，故忽略掉。
    if (c != EOF)
        ungetch(c); //将非数字字符保存起来，下次判断，如果是操作符则返回，并做相关运算，否则出错了。
    return NUMBER;
}
/*getch和ungetch函数的配对使用是为了解决去除操作数中含有操作符的问题，
因为当收集操作数时，结尾的符号是一个非数字(因为遇到非数字循环才结束)，可以用ungetch先存起来，等到下一次操作再用getch读出，*/
#define BUFSIZE 100
char buf[BUFSIZE]; /* buffer for ungetch */
int bufp = 0; /* next free position in buf */
int getch(void) /* get a (possibly pushedback)character */
{
    return (bufp > 0) ? buf[--bufp]: getchar();
}
void ungetch(int c) /* push character back on input */
{
    if (bufp >= BUFSIZE)
        printf("ungetch: too many characters\n");
    else
        buf[bufp++] = c;
}

运行效果示例：