关于strok函数的传参问题

御笔剑客

为什么第二次以后strokk函数的第一个参数要传NULL呢？这样不就没法分割指定字符串了吗？

人造人

1. /* Copyright (C) 1991 Free Software Foundation, Inc.
   
2. This file is part of the GNU C Library.
   
3. 
   
4. The GNU C Library is free software; you can redistribute it and/or
   
5. modify it under the terms of the GNU Library General Public License as
   
6. published by the Free Software Foundation; either version 2 of the
   
7. License, or (at your option) any later version.
   
8. 
   
9. The GNU C Library is distributed in the hope that it will be useful,
   
10. but WITHOUT ANY WARRANTY; without even the implied warranty of
    
11. MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the GNU
    
12. Library General Public License for more details.
    
13. 
    
14. You should have received a copy of the GNU Library General Public
    
15. License along with the GNU C Library; see the file COPYING.LIB.  If
    
16. not, write to the Free Software Foundation, Inc., 675 Mass Ave,
    
17. Cambridge, MA 02139, USA.  */
    
18. 
    
19. #include <ansidecl.h>
    
20. #include <errno.h>
    
21. #include <string.h>
    
22. 
    
23. 
    
24. static char *olds = NULL;
    
25. 
    
26. /* Parse S into tokens separated by characters in DELIM.
    
27.    If S is NULL, the last string strtok() was called with is
    
28.    used.  For example:
    
29.         char s[] = "-abc=-def";
    
30.         x = strtok(s, "-");                // x = "abc"
    
31.         x = strtok(NULL, "=-");                // x = "def"
    
32.         x = strtok(NULL, "=");                // x = NULL
    
33.                 // s = "abc\0-def\0"
    
34. */
    
35. char *
    
36. DEFUN(strtok, (s, delim),
    
37.       register char *s AND register CONST char *delim)
    
38. {
    
39.   char *token;
    
40. 
    
41.   if (s == NULL)
    
42.     {
    
43.       if (olds == NULL)
    
44.         {
    
45.           errno = EINVAL;
    
46.           return NULL;
    
47.         }
    
48.       else
    
49.         s = olds;
    
50.     }
    
51. 
    
52.   /* Scan leading delimiters.  */
    
53.   s += strspn(s, delim);
    
54.   if (*s == '\0')
    
55.     {
    
56.       olds = NULL;
    
57.       return NULL;
    
58.     }
    
59. 
    
60.   /* Find the end of the token.  */
    
61.   token = s;
    
62.   s = strpbrk(token, delim);
    
63.   if (s == NULL)
    
64.     /* This token finishes the string.  */
    
65.     olds = NULL;
    
66.   else
    
67.     {
    
68.       /* Terminate the token and make OLDS point past it.  */
    
69.       *s = '\0';
    
70.       olds = s + 1;
    
71.     }
    
72.   return token;
    
73. }
    

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人造人

1. #include <errno.h>
   
2. 
   
3. 
   
4. /* Return the length of the maximum initial segment
   
5. of S which contains only characters in ACCEPT.  */
   
6. size_t strspn(char *s, const char *accept)
   
7. {
   
8.         register const char *p;
   
9.         register const char *a;
   
10.         register size_t count = 0;
    
11. 
    
12.         for(p = s; *p != '\0'; ++p)
    
13.         {
    
14.                 for(a = accept; *a != '\0'; ++a)
    
15.                         if(*p == *a)
    
16.                                 break;
    
17.                 if(*a == '\0')
    
18.                         return count;
    
19.                 else
    
20.                         ++count;
    
21.         }
    
22. 
    
23.         return count;
    
24. }
    
25. 
    
26. 
    
27. void __cdecl exit(int _Code);
    
28. /* Cause an abnormal program termination with core-dump.  */
    
29. void abort(void)
    
30. {
    
31.         exit(127);
    
32. }
    
33. 
    
34. 
    
35. /* Find the first ocurrence of C in S.  */
    
36. char *strchr(const char *s, int c)
    
37. {
    
38.         const unsigned char *char_ptr;
    
39.         const unsigned long int *longword_ptr;
    
40.         unsigned long int longword, magic_bits, charmask;
    
41. 
    
42.         c = (unsigned char)c;
    
43. 
    
44.         /* Handle the first few characters by reading one character at a time.
    
45.         Do this until CHAR_PTR is aligned on a longword boundary.  */
    
46.         for(char_ptr = s; ((unsigned long int) char_ptr
    
47.                 & (sizeof(longword) - 1)) != 0;
    
48.                 ++char_ptr)
    
49.                 if(*char_ptr == c)
    
50.                         return (void *)char_ptr;
    
51.                 else if(*char_ptr == '\0')
    
52.                         return NULL;
    
53. 
    
54.         /* All these elucidatory comments refer to 4-byte longwords,
    
55.         but the theory applies equally well to 8-byte longwords.  */
    
56. 
    
57.         longword_ptr = (unsigned long int *) char_ptr;
    
58. 
    
59.         /* Bits 31, 24, 16, and 8 of this number are zero.  Call these bits
    
60.         the "holes."  Note that there is a hole just to the left of
    
61.         each byte, with an extra at the end:
    
62. 
    
63.         bits:  01111110 11111110 11111110 11111111
    
64.         bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
    
65. 
    
66.         The 1-bits make sure that carries propagate to the next 0-bit.
    
67.         The 0-bits provide holes for carries to fall into.  */
    
68.         switch(sizeof(longword))
    
69.         {
    
70.         case 4: magic_bits = 0x7efefeffL; break;
    
71.         case 8: magic_bits = (0x7efefefeL << 32) | 0xfefefeffL; break;
    
72.         default:
    
73.                 abort();
    
74.         }
    
75. 
    
76.         /* Set up a longword, each of whose bytes is C.  */
    
77.         charmask = c | (c << 8);
    
78.         charmask |= charmask << 16;
    
79.         if(sizeof(longword) > 4)
    
80.                 charmask |= charmask << 32;
    
81.         if(sizeof(longword) > 8)
    
82.                 abort();
    
83. 
    
84.         /* Instead of the traditional loop which tests each character,
    
85.         we will test a longword at a time.  The tricky part is testing
    
86.         if *any of the four* bytes in the longword in question are zero.  */
    
87.         for(;;)
    
88.         {
    
89.                 /* We tentatively exit the loop if adding MAGIC_BITS to
    
90.                 LONGWORD fails to change any of the hole bits of LONGWORD.
    
91. 
    
92.                 1) Is this safe?  Will it catch all the zero bytes?
    
93.                 Suppose there is a byte with all zeros.  Any carry bits
    
94.                 propagating from its left will fall into the hole at its
    
95.                 least significant bit and stop.  Since there will be no
    
96.                 carry from its most significant bit, the LSB of the
    
97.                 byte to the left will be unchanged, and the zero will be
    
98.                 detected.
    
99. 
    
100.                 2) Is this worthwhile?  Will it ignore everything except
     
101.                 zero bytes?  Suppose every byte of LONGWORD has a bit set
     
102.                 somewhere.  There will be a carry into bit 8.  If bit 8
     
103.                 is set, this will carry into bit 16.  If bit 8 is clear,
     
104.                 one of bits 9-15 must be set, so there will be a carry
     
105.                 into bit 16.  Similarly, there will be a carry into bit
     
106.                 24.  If one of bits 24-30 is set, there will be a carry
     
107.                 into bit 31, so all of the hole bits will be changed.
     
108. 
     
109.                 The one misfire occurs when bits 24-30 are clear and bit
     
110.                 31 is set; in this case, the hole at bit 31 is not
     
111.                 changed.  If we had access to the processor carry flag,
     
112.                 we could close this loophole by putting the fourth hole
     
113.                 at bit 32!
     
114. 
     
115.                 So it ignores everything except 128's, when they're aligned
     
116.                 properly.
     
117. 
     
118.                 3) But wait!  Aren't we looking for C as well as zero?
     
119.                 Good point.  So what we do is XOR LONGWORD with a longword,
     
120.                 each of whose bytes is C.  This turns each byte that is C
     
121.                 into a zero.  */
     
122. 
     
123.                 longword = *longword_ptr++;
     
124. 
     
125.                 /* Add MAGIC_BITS to LONGWORD.  */
     
126.                 if((((longword + magic_bits)
     
127. 
     
128.                         /* Set those bits that were unchanged by the addition.  */
     
129.                         ^ ~longword)
     
130. 
     
131.                         /* Look at only the hole bits.  If any of the hole bits
     
132.                         are unchanged, most likely one of the bytes was a
     
133.                         zero.  */
     
134.                         & ~magic_bits) != 0 ||
     
135. 
     
136.                         /* That caught zeroes.  Now test for C.  */
     
137.                         ((((longword ^ charmask) + magic_bits) ^ ~(longword ^ charmask))
     
138.                                 & ~magic_bits) != 0)
     
139.                 {
     
140.                         /* Which of the bytes was C or zero?
     
141.                         If none of them were, it was a misfire; continue the search.  */
     
142. 
     
143.                         const unsigned char *cp = (const unsigned char *) (longword_ptr - 1);
     
144. 
     
145.                         if(*cp == c)
     
146.                                 return (char *)cp;
     
147.                         else if(*cp == '\0')
     
148.                                 return NULL;
     
149.                         if(*++cp == c)
     
150.                                 return (char *)cp;
     
151.                         else if(*cp == '\0')
     
152.                                 return NULL;
     
153.                         if(*++cp == c)
     
154.                                 return (char *)cp;
     
155.                         else if(*cp == '\0')
     
156.                                 return NULL;
     
157.                         if(*++cp == c)
     
158.                                 return (char *)cp;
     
159.                         else if(*cp == '\0')
     
160.                                 return NULL;
     
161.                         if(sizeof(longword) > 4)
     
162.                         {
     
163.                                 if(*++cp == c)
     
164.                                         return (char *)cp;
     
165.                                 else if(*cp == '\0')
     
166.                                         return NULL;
     
167.                                 if(*++cp == c)
     
168.                                         return (char *)cp;
     
169.                                 else if(*cp == '\0')
     
170.                                         return NULL;
     
171.                                 if(*++cp == c)
     
172.                                         return (char *)cp;
     
173.                                 else if(*cp == '\0')
     
174.                                         return NULL;
     
175.                                 if(*++cp == c)
     
176.                                         return (char *)cp;
     
177.                                 else if(*cp == '\0')
     
178.                                         return NULL;
     
179.                         }
     
180.                 }
     
181.         }
     
182. 
     
183.         return NULL;
     
184. }
     
185. 
     
186. 
     
187. 
     
188. /* Find the first ocurrence in S of any character in ACCEPT.  */
     
189. char *strpbrk(register const char *s, register const char *accept)
     
190. {
     
191.         while(*s != '\0')
     
192.                 if(strchr(accept, *s) == NULL)
     
193.                         ++s;
     
194.                 else
     
195.                         return (char *)s;
     
196. 
     
197.         return NULL;
     
198. }
     
199. 
     
200. 
     
201. 
     
202. 
     
203. static char *olds = NULL;
     
204. 
     
205. /* Parse S into tokens separated by characters in DELIM.
     
206. If S is NULL, the last string strtok() was called with is
     
207. used.  For example:
     
208. char s[] = "-abc=-def";
     
209. x = strtok(s, "-");                // x = "abc"
     
210. x = strtok(NULL, "=-");                // x = "def"
     
211. x = strtok(NULL, "=");                // x = NULL
     
212. // s = "abc\0-def\0"
     
213. */
     
214. 
     
215. char *strtok(register char *s, register const char *delim)
     
216. {
     
217.         char *token;
     
218. 
     
219.         if(s == NULL)
     
220.         {
     
221.                 if(olds == NULL)
     
222.                 {
     
223.                         errno = EINVAL;
     
224.                         return NULL;
     
225.                 }
     
226.                 else
     
227.                         s = olds;
     
228.         }
     
229. 
     
230.         /* Scan leading delimiters.  */
     
231.         s += strspn(s, delim);
     
232.         if(*s == '\0')
     
233.         {
     
234.                 olds = NULL;
     
235.                 return NULL;
     
236.         }
     
237. 
     
238.         /* Find the end of the token.  */
     
239.         token = s;
     
240.         s = strpbrk(token, delim);
     
241.         if(s == NULL)
     
242.                 /* This token finishes the string.  */
     
243.                 olds = NULL;
     
244.         else
     
245.         {
     
246.                 /* Terminate the token and make OLDS point past it.  */
     
247.                 *s = '\0';
     
248.                 olds = s + 1;
     
249.         }
     
250.         return token;
     
251. }
     
252. 
     
253. #include <stdio.h>
     
254. 
     
255. int main(void)
     
256. {
     
257.         char buf[] = "hello@boy@this@is@heima";
     
258.         char *temp = strtok(buf, "@");
     
259. 
     
260.         while(temp)
     
261.         {
     
262.                 printf("%s ", temp);
     
263.                 temp = strtok(NULL, "@");
     
264.         }
     
265. 
     
266.         return 0;
     
267. }
     

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人造人

1. hello boy this is heima 请按任意键继续. . .

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