The C Programming Language 练习题4-6

题目
给计算器程序增加处理变量的命令（提供26 个具有单个英文字母变量名的变量很容易）。增加一个变量存放最近打印的值。

题目分析
由于判断条件太多，导致程序非常复杂，实现二次赋值有些麻烦，终于还是想明白了。中间加了很多打印便于分析，就不去除了，下一题还要用：）

代码实现

#include <stdio.h>#include <stdlib.h>  /* for atof() */#include <math.h>#include <string.h>#include <ctype.h>#define MAXOP 100#define NUMBER '0'#define MAXVAL 100 /* maximum depth of val stack */#define NAME 'a'int getop(char[]);void push(double);double pop(void);void mathfnc(char s[]);double varname[26];double varval[26];double val[MAXVAL]; /* value stack */int sp = 0; /* next free stack position */int main(){    int type, lastinput, i, m, n;    double op2, vtemp;    char s[MAXOP];    vtemp = 0.0;    for (i = 0; i < 26; i++)        {            varval[i] = 0.0;            varname[i] = 'a';        }    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 '%':            push((int)pop() % (int)pop());            break;        case '\n':            if(lastinput == '=')                {                    printf("\t%.8g\n", val[0]);                    vtemp = val[0];                }            else                {                vtemp = pop();                printf("\t%.8g\n", vtemp);                }            break;        case NAME:            mathfnc(s);            break;        case '=':            if(isupper(lastinput))                {                    if (sp > 1)                        pop();                    m = pop();                    n = lastinput - 'A';                    varval[n] = m;                    printf("varval[%d]=%f\tm=%d\tval[%d]=%f\n", n, varval[n], m, sp, val[sp]);                }            else                printf("error: no var\n");            break;        default:            if(isupper(type) && (varname[type - 'A'] == 'a'))/* 如果变量没有被赋值 */                varname[type - 'A'] = type;            else if(isupper(type) && (varname[type - 'A'] != 'a'))/* 如果变量已经被赋值 */                push(varval[type - 'A']);            else if (type == 'v')                push(vtemp);            else                printf("error: unknown command %s\n", s);            break;        }        lastinput = type;    }    return 0;}void mathfnc(char s[]){    double op4;    if(!strcmp(s, "sin"))        {            op4 = sin(pop());            push(op4);        }    else if (!strcmp(s, "exp"))        {            op4 = exp(pop());            push(op4);        }    else if(!strcmp(s, "pow"))        {            op4 = pop();            op4 = pow(pop(), op4);            push(op4);        }    else        printf("error: command unknown.");}/*push: push f onto value stack */void push(double f){    if (sp < MAXVAL)        {            val[sp++] = f;            printf("val[%d]=%f\t\n", 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, d;    while ((s[0] = c = getch()) == ' ' || c == '\t')        ;    s[1] = '\0';    if (!isdigit(c) && c != '.' && c != '-' && !islower(c))        return c;   /*not a number */    if ((c == '-' || c == '+') && !isdigit(d = getch()))        {            ungetch(d);            return c;        }    i = 0;    if(islower(c))        {            if (c == 'v')            return c;        {        while(islower(s[++i] = c = getch()))            ;        s[i] = '\0';        if(c != EOF)            ungetch(c);        return NAME;        }        }    if (isdigit(c))/* a number without '+' and '-' */        {            while(isdigit(s[++i] = c = getch()))        ;        }        else /* a number with '+' and '-' */        {            s[++i] = d;            while (isdigit(s[++i] = c = getch()))                ;        }    if (c == '.')   /*collect fraction part */        while (isdigit(s[++i] = c = getch()))        ;    s[i] = '\0';    if (c != EOF)        ungetch(c);    return NUMBER;}#define BUFSIZE 1000char buf[BUFSIZE];  /*buffer for ungetch */int bufp = 0;       /*next free position in buf */int getch(void)     /*get a (possibly pushed-back 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;}