gdi: Implement implicit resolving in BiDi.
This commit is contained in:
parent
032543e54d
commit
588c5a90fd
|
@ -770,6 +770,50 @@ static void resolveNeutrals(int baselevel, WORD *pcls, const WORD *plevel, int c
|
|||
SetDeferredRun(pcls, cchRun, ich, clsRun);
|
||||
}
|
||||
|
||||
/* RESOLVE IMPLICIT */
|
||||
|
||||
/*------------------------------------------------------------------------
|
||||
Function: resolveImplicit
|
||||
|
||||
Recursively resolves implicit embedding levels.
|
||||
Implements rules I1 and I2 of the Unicode Bidirectional Algorithm.
|
||||
|
||||
Input: Array of direction classes
|
||||
Character count
|
||||
Base level
|
||||
|
||||
In/Out: Array of embedding levels
|
||||
|
||||
Note: levels may exceed 15 on output.
|
||||
Accepted subset of direction classes
|
||||
R, L, AN, EN
|
||||
------------------------------------------------------------------------*/
|
||||
static const WORD addLevel[][4] =
|
||||
{
|
||||
/* L, R, AN, EN */
|
||||
/* even */ { 0, 1, 2, 2, },
|
||||
/* odd */ { 1, 0, 1, 1, }
|
||||
|
||||
};
|
||||
|
||||
static void resolveImplicit(const WORD * pcls, WORD *plevel, int cch)
|
||||
{
|
||||
int ich = 0;
|
||||
for (; ich < cch; ich++)
|
||||
{
|
||||
/* cannot resolve bn here, since some bn were resolved to strong
|
||||
* types in resolveWeak. To remove these we need the original
|
||||
* types, which are available again in resolveWhiteSpace */
|
||||
if (pcls[ich] == BN)
|
||||
{
|
||||
continue;
|
||||
}
|
||||
ASSERT(pcls[ich] > 0); /* "No Neutrals allowed to survive here." */
|
||||
ASSERT(pcls[ich] < 5); /* "Out of range." */
|
||||
plevel[ich] += addLevel[odd(plevel[ich])][pcls[ich] - 1];
|
||||
}
|
||||
}
|
||||
|
||||
/*************************************************************
|
||||
* BIDI_Reorder
|
||||
*/
|
||||
|
@ -868,6 +912,9 @@ BOOL BIDI_Reorder(
|
|||
/* resolve neutrals */
|
||||
resolveNeutrals(baselevel, chartype, levels, i);
|
||||
|
||||
/* resolveImplicit */
|
||||
resolveImplicit(chartype, levels, i);
|
||||
|
||||
/* Temporary stub: Just reverse the odd levels */
|
||||
for (j = lastgood = 0; j < i; ++j)
|
||||
if (levels[j] != levels[lastgood])
|
||||
|
|
Loading…
Reference in New Issue