#include <config.h>
#include <string.h>

Include dependency graph for strchrnul.c:

Functions
char *	strchrnul (const char *s, int c_in)
	Searching in a string. More...

Function Documentation

◆ strchrnul()

char* strchrnul	(	const char *	s,
		int	c_in
	)

Searching in a string.

This program is free software: you can redistribute it and/or modify it under the terms of the GNU General Public License as published by the Free Software Foundation; either version 3 of the License, or (at your option) any later version.

This program 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 General Public License for more details.

You should have received a copy of the GNU General Public License along with this program. If not, see http://www.gnu.org/licenses/.
Specification.
Find the first occurrence of C in S or the final NUL byte.

On 32-bit hardware, choosing longword to be a 32-bit unsigned long instead of a 64-bit uintmax_t tends to give better performance. On 64-bit hardware, unsigned long is generally 64 bits already. Change this typedef to experiment with performance.

Handle the first few bytes by reading one byte at a time. Do this until CHAR_PTR is aligned on a longword boundary.

All these elucidatory comments refer to 4-byte longwords, but the theory applies equally well to any size longwords.

Compute auxiliary longword values: repeated_one is a value which has a 1 in every byte. repeated_c has c in every byte.

Instead of the traditional loop which tests each byte, 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 equal to NUL or c. We first use an xor with repeated_c. This reduces the task to testing whether any of the four bytes in longword1 or longword2 is zero.

Let's consider longword1. We compute tmp = ((longword1 - repeated_one) & ~longword1) & (repeated_one << 7). That is, we perform the following operations:

Subtract repeated_one.
& ~longword1.
& a mask consisting of 0x80 in every byte. Consider what happens in each byte:

If a byte of longword1 is zero, step 1 and 2 transform it into 0xff, and step 3 transforms it into 0x80. A carry can also be propagated to more significant bytes.
If a byte of longword1 is nonzero, let its lowest 1 bit be at position k (0 <= k <= 7); so the lowest k bits are 0. After step 1, the byte ends in a single bit of value 0 and k bits of value 1. After step 2, the result is just k bits of value 1: 2^k - 1. After step 3, the result is 0. And no carry is produced. So, if longword1 has only non-zero bytes, tmp is zero. Whereas if longword1 has a zero byte, call j the position of the least significant zero byte. Then the result has a zero at positions 0, ..., j-1 and a 0x80 at position j. We cannot predict the result at the more significant bytes (positions j+1..3), but it does not matter since we already have a non-zero bit at position 8*j+7.

The test whether any byte in longword1 or longword2 is zero is equivalent to testing whether tmp1 is nonzero or tmp2 is nonzero. We can combine this into a single test, whether (tmp1 | tmp2) is nonzero.

This test can read more than one byte beyond the end of a string, depending on where the terminating NUL is encountered. However, this is considered safe since the initialization phase ensured that the read will be aligned, therefore, the read will not cross page boundaries and will not cause a fault.

At this point, we know that one of the sizeof (longword) bytes starting at char_ptr is == 0 or == c. On little-endian machines, we could determine the first such byte without any further memory accesses, just by looking at the tmp result from the last loop iteration. But this does not work on big-endian machines. Choose code that works in both cases.

Definition at line 24 of file strchrnul.c.

 {
   /* On 32-bit hardware, choosing longword to be a 32-bit unsigned
      long instead of a 64-bit uintmax_t tends to give better
      performance.  On 64-bit hardware, unsigned long is generally 64
      bits already.  Change this typedef to experiment with
      performance.  */
   typedef unsigned long int longword;
  
   const unsigned char *char_ptr;
   const longword *longword_ptr;
   longword repeated_one;
   longword repeated_c;
   unsigned char c;
  
   c = (unsigned char) c_in;
   if (!c)
     return rawmemchr (s, 0);
  
   /* Handle the first few bytes by reading one byte at a time.
      Do this until CHAR_PTR is aligned on a longword boundary.  */
   for (char_ptr = (const unsigned char *) s;
        (size_t) char_ptr % sizeof (longword) != 0;
        ++char_ptr)
     if (!*char_ptr || *char_ptr == c)
       return (char *) char_ptr;
  
   longword_ptr = (const longword *) char_ptr;
  
   /* All these elucidatory comments refer to 4-byte longwords,
      but the theory applies equally well to any size longwords.  */
  
   /* Compute auxiliary longword values:
      repeated_one is a value which has a 1 in every byte.
      repeated_c has c in every byte.  */
   repeated_one = 0x01010101;
   repeated_c = c | (c << 8);
   repeated_c |= repeated_c << 16;
   if (0xffffffffU < (longword) -1)
     {
       repeated_one |= repeated_one << 31 << 1;
       repeated_c |= repeated_c << 31 << 1;
       if (8 < sizeof (longword))
         {
           size_t i;
  
           for (i = 64; i < sizeof (longword) * 8; i *= 2)
             {
               repeated_one |= repeated_one << i;
               repeated_c |= repeated_c << i;
             }
         }
     }
  
   /* Instead of the traditional loop which tests each byte, 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 equal to NUL or
      c.  We first use an xor with repeated_c.  This reduces the task
      to testing whether *any of the four* bytes in longword1 or
      longword2 is zero.
  
      Let's consider longword1.  We compute tmp =
        ((longword1 - repeated_one) & ~longword1) & (repeated_one << 7).
      That is, we perform the following operations:
        1. Subtract repeated_one.
        2. & ~longword1.
        3. & a mask consisting of 0x80 in every byte.
      Consider what happens in each byte:
        - If a byte of longword1 is zero, step 1 and 2 transform it into 0xff,
          and step 3 transforms it into 0x80.  A carry can also be propagated
          to more significant bytes.
        - If a byte of longword1 is nonzero, let its lowest 1 bit be at
          position k (0 <= k <= 7); so the lowest k bits are 0.  After step 1,
          the byte ends in a single bit of value 0 and k bits of value 1.
          After step 2, the result is just k bits of value 1: 2^k - 1.  After
          step 3, the result is 0.  And no carry is produced.
      So, if longword1 has only non-zero bytes, tmp is zero.
      Whereas if longword1 has a zero byte, call j the position of the least
      significant zero byte.  Then the result has a zero at positions 0, ...,
      j-1 and a 0x80 at position j.  We cannot predict the result at the more
      significant bytes (positions j+1..3), but it does not matter since we
      already have a non-zero bit at position 8*j+7.
  
      The test whether any byte in longword1 or longword2 is zero is equivalent
      to testing whether tmp1 is nonzero or tmp2 is nonzero.  We can combine
      this into a single test, whether (tmp1 | tmp2) is nonzero.
  
      This test can read more than one byte beyond the end of a string,
      depending on where the terminating NUL is encountered.  However,
      this is considered safe since the initialization phase ensured
      that the read will be aligned, therefore, the read will not cross
      page boundaries and will not cause a fault.  */
  
   while (1)
     {
       longword longword1 = *longword_ptr ^ repeated_c;
       longword longword2 = *longword_ptr;
  
       if (((((longword1 - repeated_one) & ~longword1)
             | ((longword2 - repeated_one) & ~longword2))
            & (repeated_one << 7)) != 0)
         break;
       longword_ptr++;
     }
  
   char_ptr = (const unsigned char *) longword_ptr;
  
   /* At this point, we know that one of the sizeof (longword) bytes
      starting at char_ptr is == 0 or == c.  On little-endian machines,
      we could determine the first such byte without any further memory
      accesses, just by looking at the tmp result from the last loop
      iteration.  But this does not work on big-endian machines.
      Choose code that works in both cases.  */
  
   char_ptr = (unsigned char *) longword_ptr;
   while (*char_ptr && (*char_ptr != c))
     char_ptr++;
   return (char *) char_ptr;
 }

References c_in, and rawmemchr().

Referenced by retrieve_a_missing_file_using_external_resolver().

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Functions

Function Documentation

◆ strchrnul()