《The C programming Language》中的一个例子—逆波兰计算器问题求解
这是教材里的代码:#include<stdio.h>
#include<stdlib.h>
#include<ctype.h>
#define MAXOP 100
#define NUMBER '0'
#define MAXVAL 100
#define BUFSIZE 100
int getop(char[]);
void push(double);
double pop(void);
int getch(void);
void ungetch(int);
int sp = 0;
double val[MAXVAL];
char buf[BUFSIZE];
int bufp = 0;
void 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);
}
}
}
void push(double f)
{
if (sp < MAXVAL)
val[sp++] = f;
else
printf("error:stack full,can't push %g\n", f);
}
double pop(void)
{
if (sp > 0)
return val[--sp];
else
{
printf("error:stack empty\n");
return 0.0;
}
}
int getop(char s[])
{
int i, c;
while ((s[0] = c = getch()) == ' ' || c == '\t')
;
s[1] = '\0';
if (!isdigit(c) && c != '.')
return c;
i = 0;
if (isdigit(c))
while (isdigit(s[++i] = c = getch()))
;
if (c == '.')
while (isdigit(s[++i] = c = getch()))
;
s[i] = '\0';
if (c != EOF)
ungetch(c);
return NUMBER;
}
int getch(void)
{
return (bufp > 0) ? buf[--bufp] : getchar();
}
void ungetch(int c)
{
if (bufp >= BUFSIZE)
printf("ungetch:too many characters\n");
else
buf[bufp++] = c;
}
问题1:请帮我解释一下getop函数 81行为什么是s[1]='\0'?
问题2:getop函数最后(截图93、94、95行)为什么这样写?
[此贴子已经被作者于2016-7-4 20:26编辑过]