只匹配数字的程序？

777

1. /* this program forms the basis on which exercises 4-3 through 4-10 build */
   
2. 
   
3. #include <stdio.h>
   
4. #include <stdlib.h> /* for atof() - in K&R, math.h is referenced - this is an anachronism */
   
5. 
   
6. #define MAXOP 100 /* max size of operand or operator */
   
7. #define NUMBER '0' /* signal that a number was found */
   
8. 
   
9. int getop(char []);
   
10. void push(double);
    
11. double pop(void);
    
12. 
    
13. /* reverse Polish calculator */
    
14. 
    
15. int main(void)
    
16. {
    
17.   int type;
    
18.   double op2;
    
19.   char s[MAXOP];
    
20. 
    
21.   while((type = getop(s)) != EOF)
    
22.   {
    
23.     switch(type)
    
24.     {
    
25.       case NUMBER:
    
26.         push(atof(s));
    
27.         break;
    
28.       case '+':
    
29.         push(pop() + pop());
    
30.         break;
    
31.       case '*':
    
32.         push(pop() * pop());
    
33.         break;
    
34.       case '-':
    
35.         op2 = pop();
    
36.         push(pop() - op2);
    
37.         break;
    
38.       case '/':
    
39.         op2 = pop();
    
40.         if(op2 != 0.0)
    
41.           push(pop() / op2);
    
42.         else
    
43.           printf("error: zero divisor\n");
    
44.         break;
    
45.       case '\n':
    
46.         printf("\t%.8g\n", pop());
    
47.         break;
    
48.       default:
    
49.         printf("error: unknown command %s\n", s);
    
50.         break;
    
51.     }
    
52.   }
    
53. 
    
54.   return 0;
    
55. }
    
56. 
    
57. #define MAXVAL  100 /* maximum depth of val stack */
    
58. 
    
59. int sp = 0; /* next free stack position */
    
60. double val[MAXVAL]; /* value stack */
    
61. 
    
62. /* push: push f onto value stack */
    
63. void push(double f)
    
64. {
    
65.   if(sp < MAXVAL)
    
66.     val[sp++] = f;
    
67.   else
    
68.     printf("error: stack full, can't push %g\n", f);
    
69. }
    
70. 
    
71. /* pop: pop and return top value from stack */
    
72. double pop(void)
    
73. {
    
74.   if(sp > 0)
    
75.     return val[--sp];
    
76.   else
    
77.   {
    
78.     printf("error: stack empty\n");
    
79.     return 0.0;
    
80.   }
    
81. }
    
82. 
    
83. #include <ctype.h>
    
84. 
    
85. int getch(void);
    
86. void ungetch(int);
    
87. 
    
88. /* getop: get next operator or numeric operand */
    
89. int getop(char s[])
    
90. {
    
91.   int i, c;
    
92. 
    
93.   while((s[0] = c = getch()) == ' ' || c == '\t')
    
94.     ;
    
95. 
    
96.   s[1] = '\0';
    
97.   if(!isdigit(c) && c != '.')
    
98.     return c; /* not a number */
    
99.   i = 0;
    
100.   if(isdigit(c)) /* collect integer part */
     
101.     while(isdigit(s[++i] = c = getch()))
     
102.       ;
     
103.   if(c == '.')
     
104.     while(isdigit(s[++i] = c = getch()))
     
105.       ;
     
106.   s[i] = '\0';
     
107.   if(c != EOF)
     
108.     ungetch(c);
     
109.   return NUMBER;
     
110. }
     
111. 
     
112. #define BUFSIZE 100
     
113. 
     
114. char buf[BUFSIZE]; /* buffer for ungetch */
     
115. int bufp = 0; /* next free position in buf */
     
116. 
     
117. int getch(void) /* get a (possibly pushed back) character */
     
118. {
     
119.   return (bufp > 0) ? buf[--bufp] : getchar();
     
120. }
     
121. 
     
122. void ungetch(int c) /* push character back on input */
     
123. {
     
124.   if(bufp >= BUFSIZE)
     
125.     printf("ungetch: too many characters\n");
     
126.   else
     
127.     buf[bufp++] = c;
     
128. }
     

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不知道这个程序是干嘛的？

kj501

如果学过编译原理就清楚啦。这就是利用逆波兰表达式实现计算器。如果表达式为2+3*5，转化成逆波兰表达式输入就是3 5 * 2 +，再利用堆栈，先压入3和5，碰到*时弹出3和5计算乘积得15并压入堆栈得15，再压入2，碰到+时再弹出15和2，得出其和为17。