item.select()

item.get_submenu().popup(menubar, item, move_menu, 0, 0)

item.get_submenu().popup(None, None, move_menu, 0, 0)

#include <string.h>
#include <stdlib.h>

#undef strlen

/* Return the length of the null-terminated string STR.  Scan for
   the null terminator quickly by testing four bytes at a time.  */
size_t
strlen (str)
     const char *str;
{
  const char *char_ptr;
  const unsigned long int *longword_ptr;
  unsigned long int longword, himagic, lomagic;

  /* 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 = str; ((unsigned long int) char_ptr
            & (sizeof (longword) - 1)) != 0;
       ++char_ptr)
    if (*char_ptr == '\0')
      return char_ptr - str;

  /* 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.  */
  himagic = 0x80808080L;
  lomagic = 0x01010101L;
  if (sizeof (longword) > 4)
    {
      /* 64-bit version of the magic.  */
      /* Do the shift in two steps to avoid a warning if long has 32 bits.  */
      himagic = ((himagic << 16) << 16) | himagic;
      lomagic = ((lomagic << 16) << 16) | lomagic;
    }
  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 (;;)
    {
      longword = *longword_ptr++;

      if (((longword - lomagic) & ~longword & himagic) != 0)
    {
      /* Which of the bytes was the zero?  If none of them were, it was
         a misfire; continue the search.  */

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

      if (cp[0] == 0)
        return cp - str;
      if (cp[1] == 0)
        return cp - str + 1;
      if (cp[2] == 0)
        return cp - str + 2;
      if (cp[3] == 0)
        return cp - str + 3;
      if (sizeof (longword) > 4)
        {
          if (cp[4] == 0)
        return cp - str + 4;
          if (cp[5] == 0)
        return cp - str + 5;
          if (cp[6] == 0)
        return cp - str + 6;
          if (cp[7] == 0)
        return cp - str + 7;
        }
    }
    }
}
libc_hidden_builtin_def (strlen)

#include <sys/cdefs.h>
#include <string.h>

size_t
strlen(str)
    const char *str;
{
    register const char *s;

    for (s = str; *s; ++s);
    return(s - str);
}

#include <sys/cdefs.h>
__FBSDID("$FreeBSD: src/lib/libc/string/strlen.c,v 1.10 2010/03/13 00:15:06 delphij Exp $");

#include <sys/limits.h>
#include <sys/types.h>
#include <string.h>

/*
 * Portable strlen() for 32-bit and 64-bit systems.
 *
 * Rationale: it is generally much more efficient to do word length
 * operations and avoid branches on modern computer systems, as
 * compared to byte-length operations with a lot of branches.
 *
 * The expression:
 *
 *    ((x - 0x01....01) & ~x & 0x80....80)
 *
 * would evaluate to a non-zero value iff any of the bytes in the
 * original word is zero.
 *
 * On multi-issue processors, we can divide the above expression into:
 *    a)  (x - 0x01....01)
 *    b) (~x & 0x80....80)
 *    c) a & b
 *
 * Where, a) and b) can be partially computed in parallel.
 *
 * The algorithm above is found on "Hacker's Delight" by
 * Henry S. Warren, Jr.
 */

/* Magic numbers for the algorithm */
#if LONG_BIT == 32
static const unsigned long mask01 = 0x01010101;
static const unsigned long mask80 = 0x80808080;
#elif LONG_BIT == 64
static const unsigned long mask01 = 0x0101010101010101;
static const unsigned long mask80 = 0x8080808080808080;
#else
#error Unsupported word size
#endif

#define    LONGPTR_MASK (sizeof(long) - 1)

/*
 * Helper macro to return string length if we caught the zero
 * byte.
 */
#define testbyte(x)                \
    do {                    \
        if (p[x] == '\0')        \
            return (p - str + x);    \
    } while (0)

size_t
strlen(const char *str)
{
    const char *p;
    const unsigned long *lp;
    long va, vb;

    /*
     * Before trying the hard (unaligned byte-by-byte access) way
     * to figure out whether there is a nul character, try to see
     * if there is a nul character is within this accessible word
     * first.
     *
     * p and (p & ~LONGPTR_MASK) must be equally accessible since
     * they always fall in the same memory page, as long as page
     * boundaries is integral multiple of word size.
     */
    lp = (const unsigned long *)((uintptr_t)str & ~LONGPTR_MASK);
    va = (*lp - mask01);
    vb = ((~*lp) & mask80);
    lp++;
    if (va & vb)
        /* Check if we have \0 in the first part */
        for (p = str; p < (const char *)lp; p++)
            if (*p == '\0')
                return (p - str);

    /* Scan the rest of the string using word sized operation */
    for (; ; lp++) {
        va = (*lp - mask01);
        vb = ((~*lp) & mask80);
        if (va & vb) {
            p = (const char *)(lp);
            testbyte(0);
            testbyte(1);
            testbyte(2);
            testbyte(3);
#if (LONG_BIT >= 64)
            testbyte(4);
            testbyte(5);
            testbyte(6);
            testbyte(7);
#endif
        }
    }

    /* NOTREACHED */
    return (0);
}

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2) 31    cm-t
3) 26    Кຼزດ
4) 24    nesthib
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7) 13    na kraïou
8) 9    Rolinh
9) 5    grim7reaper
9) 5    :!pakman
11) 4    \\Ouranos//
11) 4    The Uploader
11) 4    kamui57

1) 3428    nesthib
2) 3065    samuncle
3) 2893    Pylade
4) 2178    Кຼزດ
5) 1698+5  grim7reaper /* ./viewtopic.php?pid=3486252#p3486252 */
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9) 862    \\Ouranos//
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