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关于strok函数的传参问题

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发表于 2018-4-30 13:22:27 | 显示全部楼层 |阅读模式

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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?
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发表于 2018-4-30 19:44:50 | 显示全部楼层
/* Copyright (C) 1991 Free Software Foundation, Inc.
This file is part of the GNU C Library.

The GNU C Library is free software; you can redistribute it and/or
modify it under the terms of the GNU Library General Public License as
published by the Free Software Foundation; either version 2 of the
License, or (at your option) any later version.

The GNU C Library is distributed in the hope that it will be useful,
but WITHOUT ANY WARRANTY; without even the implied warranty of
MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the GNU
Library General Public License for more details.

You should have received a copy of the GNU Library General Public
License along with the GNU C Library; see the file COPYING.LIB.  If
not, write to the Free Software Foundation, Inc., 675 Mass Ave,
Cambridge, MA 02139, USA.  */

#include <ansidecl.h>
#include <errno.h>
#include <string.h>


static char *olds = NULL;

/* Parse S into tokens separated by characters in DELIM.
   If S is NULL, the last string strtok() was called with is
   used.  For example:
        char s[] = "-abc=-def";
        x = strtok(s, "-");                // x = "abc"
        x = strtok(NULL, "=-");                // x = "def"
        x = strtok(NULL, "=");                // x = NULL
                // s = "abc\0-def\0"
*/
char *
DEFUN(strtok, (s, delim),
      register char *s AND register CONST char *delim)
{
  char *token;

  if (s == NULL)
    {
      if (olds == NULL)
        {
          errno = EINVAL;
          return NULL;
        }
      else
        s = olds;
    }

  /* Scan leading delimiters.  */
  s += strspn(s, delim);
  if (*s == '\0')
    {
      olds = NULL;
      return NULL;
    }

  /* Find the end of the token.  */
  token = s;
  s = strpbrk(token, delim);
  if (s == NULL)
    /* This token finishes the string.  */
    olds = NULL;
  else
    {
      /* Terminate the token and make OLDS point past it.  */
      *s = '\0';
      olds = s + 1;
    }
  return token;
}
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发表于 2018-4-30 20:33:31 | 显示全部楼层
#include <errno.h>


/* Return the length of the maximum initial segment
of S which contains only characters in ACCEPT.  */
size_t strspn(char *s, const char *accept)
{
        register const char *p;
        register const char *a;
        register size_t count = 0;

        for(p = s; *p != '\0'; ++p)
        {
                for(a = accept; *a != '\0'; ++a)
                        if(*p == *a)
                                break;
                if(*a == '\0')
                        return count;
                else
                        ++count;
        }

        return count;
}


void __cdecl exit(int _Code);
/* Cause an abnormal program termination with core-dump.  */
void abort(void)
{
        exit(127);
}


/* Find the first ocurrence of C in S.  */
char *strchr(const char *s, int c)
{
        const unsigned char *char_ptr;
        const unsigned long int *longword_ptr;
        unsigned long int longword, magic_bits, charmask;

        c = (unsigned char)c;

        /* Handle the first few characters by reading one character at a time.
        Do this until CHAR_PTR is aligned on a longword boundary.  */
        for(char_ptr = s; ((unsigned long int) char_ptr
                & (sizeof(longword) - 1)) != 0;
                ++char_ptr)
                if(*char_ptr == c)
                        return (void *)char_ptr;
                else if(*char_ptr == '\0')
                        return NULL;

        /* All these elucidatory comments refer to 4-byte longwords,
        but the theory applies equally well to 8-byte longwords.  */

        longword_ptr = (unsigned long int *) char_ptr;

        /* Bits 31, 24, 16, and 8 of this number are zero.  Call these bits
        the "holes."  Note that there is a hole just to the left of
        each byte, with an extra at the end:

        bits:  01111110 11111110 11111110 11111111
        bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD

        The 1-bits make sure that carries propagate to the next 0-bit.
        The 0-bits provide holes for carries to fall into.  */
        switch(sizeof(longword))
        {
        case 4: magic_bits = 0x7efefeffL; break;
        case 8: magic_bits = (0x7efefefeL << 32) | 0xfefefeffL; break;
        default:
                abort();
        }

        /* Set up a longword, each of whose bytes is C.  */
        charmask = c | (c << 8);
        charmask |= charmask << 16;
        if(sizeof(longword) > 4)
                charmask |= charmask << 32;
        if(sizeof(longword) > 8)
                abort();

        /* Instead of the traditional loop which tests each character,
        we will test a longword at a time.  The tricky part is testing
        if *any of the four* bytes in the longword in question are zero.  */
        for(;;)
        {
                /* We tentatively exit the loop if adding MAGIC_BITS to
                LONGWORD fails to change any of the hole bits of LONGWORD.

                1) Is this safe?  Will it catch all the zero bytes?
                Suppose there is a byte with all zeros.  Any carry bits
                propagating from its left will fall into the hole at its
                least significant bit and stop.  Since there will be no
                carry from its most significant bit, the LSB of the
                byte to the left will be unchanged, and the zero will be
                detected.

                2) Is this worthwhile?  Will it ignore everything except
                zero bytes?  Suppose every byte of LONGWORD has a bit set
                somewhere.  There will be a carry into bit 8.  If bit 8
                is set, this will carry into bit 16.  If bit 8 is clear,
                one of bits 9-15 must be set, so there will be a carry
                into bit 16.  Similarly, there will be a carry into bit
                24.  If one of bits 24-30 is set, there will be a carry
                into bit 31, so all of the hole bits will be changed.

                The one misfire occurs when bits 24-30 are clear and bit
                31 is set; in this case, the hole at bit 31 is not
                changed.  If we had access to the processor carry flag,
                we could close this loophole by putting the fourth hole
                at bit 32!

                So it ignores everything except 128's, when they're aligned
                properly.

                3) But wait!  Aren't we looking for C as well as zero?
                Good point.  So what we do is XOR LONGWORD with a longword,
                each of whose bytes is C.  This turns each byte that is C
                into a zero.  */

                longword = *longword_ptr++;

                /* Add MAGIC_BITS to LONGWORD.  */
                if((((longword + magic_bits)

                        /* Set those bits that were unchanged by the addition.  */
                        ^ ~longword)

                        /* Look at only the hole bits.  If any of the hole bits
                        are unchanged, most likely one of the bytes was a
                        zero.  */
                        & ~magic_bits) != 0 ||

                        /* That caught zeroes.  Now test for C.  */
                        ((((longword ^ charmask) + magic_bits) ^ ~(longword ^ charmask))
                                & ~magic_bits) != 0)
                {
                        /* Which of the bytes was C or zero?
                        If none of them were, it was a misfire; continue the search.  */

                        const unsigned char *cp = (const unsigned char *) (longword_ptr - 1);

                        if(*cp == c)
                                return (char *)cp;
                        else if(*cp == '\0')
                                return NULL;
                        if(*++cp == c)
                                return (char *)cp;
                        else if(*cp == '\0')
                                return NULL;
                        if(*++cp == c)
                                return (char *)cp;
                        else if(*cp == '\0')
                                return NULL;
                        if(*++cp == c)
                                return (char *)cp;
                        else if(*cp == '\0')
                                return NULL;
                        if(sizeof(longword) > 4)
                        {
                                if(*++cp == c)
                                        return (char *)cp;
                                else if(*cp == '\0')
                                        return NULL;
                                if(*++cp == c)
                                        return (char *)cp;
                                else if(*cp == '\0')
                                        return NULL;
                                if(*++cp == c)
                                        return (char *)cp;
                                else if(*cp == '\0')
                                        return NULL;
                                if(*++cp == c)
                                        return (char *)cp;
                                else if(*cp == '\0')
                                        return NULL;
                        }
                }
        }

        return NULL;
}



/* Find the first ocurrence in S of any character in ACCEPT.  */
char *strpbrk(register const char *s, register const char *accept)
{
        while(*s != '\0')
                if(strchr(accept, *s) == NULL)
                        ++s;
                else
                        return (char *)s;

        return NULL;
}




static char *olds = NULL;

/* Parse S into tokens separated by characters in DELIM.
If S is NULL, the last string strtok() was called with is
used.  For example:
char s[] = "-abc=-def";
x = strtok(s, "-");                // x = "abc"
x = strtok(NULL, "=-");                // x = "def"
x = strtok(NULL, "=");                // x = NULL
// s = "abc\0-def\0"
*/

char *strtok(register char *s, register const char *delim)
{
        char *token;

        if(s == NULL)
        {
                if(olds == NULL)
                {
                        errno = EINVAL;
                        return NULL;
                }
                else
                        s = olds;
        }

        /* Scan leading delimiters.  */
        s += strspn(s, delim);
        if(*s == '\0')
        {
                olds = NULL;
                return NULL;
        }

        /* Find the end of the token.  */
        token = s;
        s = strpbrk(token, delim);
        if(s == NULL)
                /* This token finishes the string.  */
                olds = NULL;
        else
        {
                /* Terminate the token and make OLDS point past it.  */
                *s = '\0';
                olds = s + 1;
        }
        return token;
}

#include <stdio.h>

int main(void)
{
        char buf[] = "hello@boy@this@is@heima";
        char *temp = strtok(buf, "@");

        while(temp)
        {
                printf("%s ", temp);
                temp = strtok(NULL, "@");
        }

        return 0;
}
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发表于 2018-4-30 20:34:22 | 显示全部楼层
hello boy this is heima 请按任意键继续. . .
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