Thank you for advice, but the algorithm would be "OR", so, considering
the proposal of Kan-Ru Chen, the unified conditional would be:
if ( (face->family_name && tt_check_trickyness_family( face->family_name )) ||
tt_check_trickyness_sfnt_ids( (TT_Face)face ) )
return TRUE;
return FALSE;
If I'm misunderstanding your proposal, please let me know.
Regards,
mpsuzuki
Алексей Подтележников wrote:
> On Mon, Apr 25, 2011 at 10:34 AM, <[email protected]> wrote:
>> so I will apply your patch.
>
> You wrapped two nested if-conditions into one with &&.
> It could be wrapped with the preceding if-condition further.
>
>
> /* Type42 fonts may lack `name' tables, we thus try to identify */
> /* tricky fonts by checking the checksums of Type42-persistent */
> /* sfnt tables (`cvt', `fpgm', and `prep'). Then check family name. */
> if ( tt_check_trickyness_sfnt_ids( (TT_Face)face ) &&
> face->family_name &&
> tt_check_trickyness_family( face->family_name ) )
> return TRUE;
>
> return FALSE;
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