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 100char 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：