t61.c (ITS#2388)

[peter.gutekunst@secrypt.de]

Full_Name: Peter Gutekunst
Version: 2.1.13
OS: Windows2000
URL: ftp://ftp.openldap.org/incoming/
Submission from: (NULL) (217.225.11.245)


When we worked with openldap-2.1.13-module t61.c we had a problem with the 

following routine when we tried to transform strings containing letters with 

diaeresesis:

/* Transform a T.61 string to UTF-8.

*/

int ldap_t61s_to_utf8s( struct berval *src, struct berval *dst )

{

  unsigned char *c;

  char *d;

  int i, wlen = 0;

  /* Just count the length of the UTF-8 result first */

  for (i=0,c=(unsigned char *)src->bv_val; i < src->bv_len; c++,i++) {

    /* Invalid T.61 characters? */

      if (!t61_tab[*c]) 

       return LDAP_INVALID_SYNTAX;

     if (*c & 0xf0 == 0xc0) {


/*

* PROBLEM!! the comparison-operator(==) was stronger then the bitwise and(&) so

* the if-block was never entered. By bracketing as follows we got better
results.

* if ((*c & 0xf0) == 0xc0) {

*/

    int j = *c & 0x0f;

    /* If this is the end of the string, or if the base

    * character is just a space, treat this as a regular

    * spacing character.

    */

    if ((!c[1] || c[1] == 0x20) && accents[j]) {

     wlen += ldap_x_wc_to_utf8(NULL, accents[j], 0);

    } else if (cx_tab[j] && cx_tab[j][c[1]>>5] &&

    /* We have a composite mapping for this pair */

                (*cx_tab[j][c[1]>>5])[c[1]&0x1f]) {

      wlen += ldap_x_wc_to_utf8( NULL,(*cx_tab[j][c[1]>>5])[c[1]&0x1f], 0);

    } else {

      /* No mapping, just swap it around so the base

      * character comes first.

      */

      wlen += ldap_x_wc_to_utf8(NULL, c[1], 0);

      wlen += ldap_x_wc_to_utf8(NULL, t61_tab[*c], 0);

    }

    c++; i++;

    continue;

    } else {

      wlen += ldap_x_wc_to_utf8(NULL, t61_tab[*c], 0);

    }

  }

  /* Now transform the string */

  dst->bv_len = wlen;

  dst->bv_val = LDAP_MALLOC( wlen+1 );

  d = dst->bv_val;

  if (!d)

    return LDAP_NO_MEMORY;

 for (i=0,c=(unsigned char *)src->bv_val; i < src->bv_len; c++,i++) {

    if (*c & 0xf0 == 0xc0) {

      /*

      * PROBLEM!! the comparison-operator(==) was stronger then the bitwise
and(&) so

      * the if-block was never entered.

     */

     int j = *c & 0x0f;

  .......

}