The reqiured arguement num of cff_op_dup defined in cff_argument_counts is 1.
Before processing cff_op_dup, agrs minus 1 to points the first arguement of
cff_op_dup. The comment said:
/* At this point, `args' points to the first argument of the */
/* operand in case `req_args' isn't zero. Otherwise, we have */
/* to adjust `args' manually. */
So, if cff_op_dup duplicates args[0], args should add 2 to keep these two
elements on stack.
According to codes of cff_op_abs, cff_op_and and cff_op_exch processing:
case cff_op_abs:
FT_TRACE4(( abs\n ));
if ( args[0] 0 )
args[0] = -args[0];
args++;
break;
case cff_op_add:
FT_TRACE4(( add\n ));
args[0] += args[1];
args++;
break;
case cff_op_exch:
{
FT_Fixed tmp;
FT_TRACE4(( exch\n ));
tmp = args[0];
args[0] = args[1];
args[1] = tmp;
args += 2;
}
break;
cff_op_abs consumes 1 and push 1, args++ is right;
cff_op_add consumes 2 and push 1, args++ is right;
cff_op_exch consumes 2 and push 2, args +=2 is right;
cff_op_dup consumes 1 and push 2, args +=2 should be right;
--
Ning Dong
2009-10-10
-
发件人：Graham Asher
发送日期：2009-10-09 22:48:53
收件人：flintning
抄送：freetype-devel
主题：Re: [ft-devel] Question about CFF arithmetic operator dup
The net effect of consuming one element and adding two is to add one.
The code looks correct to me.
Graham Asher
Ning Dong wrote:
Hi,
I'm checking the cff_decoder_parse_charstrings function in cffgload.c
of FT239.
According to CFF technical note #5177, operator dup should
consume one element and put two elements to arguement stack:
dup any dup(12 27) any any
But in cff_decoder_parse_charstrings, the codes like this:
case cff_op_dup:
FT_TRACE4(( dup\n ));
args[1] = args[0];
args++;//should be args+=2?
break;
Only one element put to arguement stack.
Is this a bug?
FT2310 is the same.
2009-10-09
Ning Dong
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