K&R可能在波兰计算器中出错

Question

我不知道在哪里发布这个,但我想我在K&R的波兰计算器程序中发现了一个相当大的错误.基本上,当您执行操作时,会弹出两个值,而只会推送结果.问题是结果没有被推到堆栈顶部!这是一个例子:

教科书提供的波兰计算器的完整代码如下所示:

#include <stdio.h>
#include <stdlib.h> /* for atof() */

#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 '\n':
            printf("\t%.8g\n", pop());
            break;
        default:
            printf("error: unknown command %s\n", s);
            break;
        }
    }
    system("Pause");
    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 operator or numeric operand */
int getop(char s[])
{
    int i, c;

    while ((s[0] = c = getch()) == ' ' || c == '\t')
        ;
    s[1] = '\0';
    if (!isdigit(c) && c != '.')
        return c; /* not a number */
    i = 0;
    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';
    if (c != EOF)
        ungetch(c);
    return NUMBER;
}

#define BUFSIZE 100

char 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;
}

如果你想自己检查,我所做的就是添加

static int pass = 0;
int i, check;
i = check = 0;

在main()和的while循环中

if(!check) {
    printf("pass #%d\n",pass++);
    while(val[i] != '\0') {
        printf("val[%d]: %.2f\n",i,val[i]);
        ++i;
    }
}

在while循环结束时,就在switch语句之后.我也提出check = 1;了这个案子'\n'.

作为一种可能的解决方法,我重新编写了pop函数,以便一旦访问它们就会从val数组中删除弹出值.而不是

double pop(void)
{
    if (sp > 0)
        return val[--sp];
    else {
        printf("error: stack empty\n");
        return 0.0;
    }
}

你有类似的东西

double pop(void)
{
    if (sp > 0) {
        double temp = val[--sp];
        val[sp] = '\0';
        return temp;
    }
    else {
        printf("error: stack empty\n");
        return 0.0;
    }
}

我还重写了push函数,以确保值始终被推送到val数组的末尾.而不是

void push(double f)
{
    if ( sp < MAXVAL)
        val[sp++] = f;
    else
        printf("error: stack full. can't push %g\n", f);
}

你有

void push(double f)
{
    if ( sp < MAXVAL) {
        while (val[sp] != '\0')
            ++sp;
        val[sp++] = f;
    }
    else
        printf("error: stack full. can't push %g\n", f);
}

即使有了这些变化,我仍然需要重新编写

case '\n':
        printf("\t%.8g\n", pop());
        break;

检索堆栈顶部的值而不弹出它,这需要用printf一个简单的函数替换语句

void print_top(void)
{
    int i = 0;
    while( val[i] != '\0' )
        ++i;
    --i;
    printf("\t%.8g\n",val[i]);
}

只有这样,抛光计算器似乎按预期运行,至少就堆栈在幕后的作用而言.您可以使用修改后的代码自行尝试:

#include <stdio.h>
#include <stdlib.h> /* for atof() */
#include <ctype.h>

#define MAXOP 100 /* max size of operand or operator */
#define NUMBER '0' /* signal that a number was found */
#define MAXVAL 100 /* maximum depth of val stack */

int getop(char []);
void push(double);
double pop(void);
void print_top(void);

int sp = 0; /* next free stack position */
double val[MAXVAL]; /* value stack */

/* reverse Polish calculator */
main()
{
    int type;
    double op2;
    char s[MAXOP];

    while ((type= getop(s)) != EOF) {

        static int pass = 0;
        int i, check;
        i = check = 0;

        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 '\n':
            print_top();
            check = 1;
            break;
        default:
            printf("error: unknown command %s\n", s);
            break;
        }
        if(!check) {
            printf("pass #%d\n",pass++);
            while(val[i] != '\0') {
                printf("val[%d]: %.2f\n",i,val[i]);
                ++i;
            }
        }
    }
    system("Pause");
    return 0;
}

/* push: push f onto value stack */
void push(double f)
{
    if ( sp < MAXVAL) {
        while (val[sp] != '\0')
            ++sp;
        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) {
        double temp = val[--sp];
        val[sp] = '\0';
        return temp;
    }
    else {
        printf("error: stack empty\n");
        return 0.0;
    }
}

int getch(void);
void ungetch(int);

/* getop: get next operator or numeric operand */
int getop(char s[])
{
    int i, c;

    while ((s[0] = c = getch()) == ' ' || c == '\t')
        ;
    s[1] = '\0';
    if (!isdigit(c) && c != '.')
        return c; /* not a number */
    i = 0;
    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';
    if (c != EOF)
        ungetch(c);
    return NUMBER;
}

#define BUFSIZE 100

char 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;
}

void print_top(void)
{
    int i = 0;
    while( val[i] != '\0' )
        ++i;
    --i;
    printf("\t%.8g\n",val[i]);
}

请注意,#define为了适应最后的调试printf语句,我必须将大部分语句和原型声明移到开头main().

*编辑了我的一些大胆的主张:P

Answer 1

1. 您正在考虑向后堆栈 - 堆栈顶部是最高有效索引,而不是val[0].当你看到操作数的推动时,这种行为是显而易见的.你的输出:

   3 4 +
   pass #0
   val[0]: 3.00
   pass #1
   val[0]: 3.00
   val[1]: 4.00
   

   首先,3推送 - 进入(先前为空)堆栈的顶部 - 它位于插槽0中.接下来4被推送.正如您所看到的,它进入了val[1],清楚地表明val[0]在这种情况下它不是堆栈的顶部.

2. 您错误地打印了堆栈,这让您感到困惑.将您的打印循环更改为:

   while (i < sp) {
       printf("val[%d]: %.2f\n",i,val[i]);
       ++i;
   }
   

   也就是说,只打印堆栈中的有效条目,您将看到错误.

   您当前的比较是0在堆栈上查找条目,而不是程序识别空闲条目的方式.这就是sp变量的用途.除了寻找错误的东西,它还是以一种奇怪的方式进行 - val是一个浮点数的数组 - 你为什么要比较一个字符文字\0？

   这是完整的更正输出:

   3 4 +
   pass #0
   val[0]: 3.00
   pass #1
   val[0]: 3.00
   val[1]: 4.00
   pass #2
   val[0]: 7.00
       7
   

   现在,您可以看到正确的输出-无论是3.00和4.00被弹出,并且7.00被推回压入堆栈.它现在是唯一有效的条目.