RE: Converting Outlines to postscript Vector

[Just van Rossum]

At 9:07 AM +0000 29-02-2000, Satpal Chander wrote:
>> I'm not 100% sure, but it seems FT1.x doesn't have the decompose outline
>> API like FT2 has, so I guess you'll have to walk trhough the
>> outline object yourself.
>
>If this is the case then how stable is the FT2 stuff considering that I'm
>only interested in the outline data for each glyph.

I can't answer that. Still: just iterating over the outline yourself is
hardly harder than using the FT_Outline_Decompose() function in FT2.

>> Even so, it should be fairly straightforward, though:
>>
>> - There are on-curve and off-curve points in TT (and therefore in FT1.x)
>> - Between any two consecutive off-curve points there is an
>> implied on-curve
>> point (exactly in the middle between them).
>> - A quadratic segment (onCurve, offCurve, onCurve) can be converted to
>> cubic beziers (as used by PS) like so:
>>    let (from, control, to) be the three points making up de
>> quadratic curve
>>    you need to convert these three points to four points for
>> cubics: (from,
>> control1, control2, to)
>>        control1 = from + (control - from) * 2.0/3.0
>>        control2 = to + (control - to) * 2.0/3.0
>>    from and to remain the same.
>
>Does the algorithm above meant that I don't have to actually
>do anything with the curves using the curve maths, just manipulating the
>control points and outputting them to postscript. ;-))

That's correct. TT's quadratic curves can be converted to PS's cubic
beziers with the above algorithm without any loss of quality. (The reverse
is *not* possible without losing information.)

Just