[cairo] How to get extents of transformed graphics
Andreas Falkenhahn
andreas at falkenhahn.com
Fri Sep 9 17:23:10 UTC 2022
On 09.09.2022 at 17:38 Bill Spitzak wrote:
> Not sure exactly what you are trying to do, but this is probably
> what you want (this also assumes x2>x1 and y2>y1):
> width = ceil(x2 * 4.6) - floor(x1 * 4.6);
> height = ceil(y2 * 4.6) - floor(y1 * 4.6);
Thanks for your answer! Unfortunately, the "C" shape still gets cut off so the surface height is still too small.
> The reason for the ceil and floor is so the fractional-covered
> pixels are rounded out to the result. You have to do this after you
> convert to pixels which is what multiplying by 4.6 is doing (?).
No, 4.6 is the scaling factor on both axes. What I'm trying to do is this: Allocate an image surface so that a transformed path fits into it exactly. I'm transforming the "C" shape by applying an affine transformation with sx=4.6 and sy=4.6. Now I need to find out the size in pixels for this transformed path so that I can allocate an image surface for it.
Here's the test program. It draws the transformed "C" shape to an image surface and then saves this image surface to a PNG. As you can see in the output PNG, a row is missing at the bottom of the image.
#include <ctype.h>
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <cairo.h>
int main(int argc, char *argv[])
{
cairo_t *cr;
cairo_surface_t *surface;
cairo_matrix_t cm;
int k;
double tx = 0, ty = 0;
double x1, y1, x2, y2;
surface = cairo_image_surface_create(CAIRO_FORMAT_A8, 1000, 1000);
cr = cairo_create(surface);
for(k = 0; k < 2; k++) {
cairo_new_path(cr);
cairo_move_to(cr, tx, ty);
cairo_set_font_size(cr, 100);
cairo_text_path(cr, "C");
if(!k) {
double width, height;
cairo_fill_extents(cr, &x1, &y1, &x2, &y2);
tx = -x1;
ty = -y1;
width = ceil(x2 * 4.6) - floor(x1 * 4.6);
height = ceil(y2 * 4.6) - floor(y1 * 4.6);
cairo_destroy(cr);
cairo_surface_destroy(surface);
surface = cairo_image_surface_create(CAIRO_FORMAT_A8, width, height);
cr = cairo_create(surface);
cairo_matrix_init(&cm, 4.6, 0, 0, 4.6, 0, 0);
cairo_set_matrix(cr, &cm);
}
}
cairo_fill(cr);
cairo_surface_flush(surface);
cairo_surface_write_to_png(surface, "tmp.png");
cairo_destroy(cr);
cairo_surface_destroy(surface);
return 0;
}
> On Fri, Sep 9, 2022 at 7:16 AM Andreas Falkenhahn <andreas at falkenhahn.com> wrote:
> I can fix the issue by adding 2 to "height" like so:
>
> width = fabs(ceil(x2) - floor(x1));
> height = fabs(ceil(y2) - floor(y1));
>
> width = ceil(width * 4.6);
> height = ceil(height * 4.6) + 2;
>
> But this feels very kludgy and probably only works for my specific
> case and not for any arbitrary transformation values.
>
> Is there really no general solution to this problem? IMHO it's a
> rather common thing to do and I'm a bit puzzled why this should be so difficult to solve...
>
> On 08.09.2022 at 19:24 Andreas Falkenhahn wrote:
>
>> I've tried this:
>
>> width = fabs(x2 - floor(x1));
>> height = fabs(y2 - floor(y1));
>
>> width = ceil(width * 4.6);
>> height = ceil(height * 4.6);
>
>> Still cuts off a line of the "C" shape. I've also tried this:
>
>> width = fabs(ceil(x2) - floor(x1));
>> height = fabs(ceil(y2) - floor(y1));
>
>> width = ceil(width * 4.6);
>> height = ceil(height * 4.6);
>
>> No success either. A line is cut off.
>
>> Any more ideas? You can find my test program below...
>
>> On 08.09.2022 at 19:03 Behdad Esfahbod wrote:
>
>>> Try flooring x1,y1 before subtracting them from x2,y2 for computing the size of your surface.
>
>
>>> behdad
>>> http://behdad.org/
>
>>> On Thu, Sep 8, 2022 at 10:21 AM Andreas Falkenhahn <andreas at falkenhahn.com> wrote:
>
>>> Is there really nobody who knows how to solve this? There surely
>>> must be a way to find out the *exact* extents of transformed
>>> graphics? The idea is to find out the exact dimensions in order to
>>> allocate a Cairo surface that *exactly* matches the size of the path.
>>>
>>> On 28.08.2022 at 21:47 Andreas Falkenhahn wrote:
>>>
>>>> Hi,
>>>
>>>> is there any way to get the *exact* extents of transformed
>>>> graphics? AFAICS, cairo_fill_extents() is in user coordinates, i.e.
>>>> it doesn't take any transformation settings into account. So I've
>>>> tried to apply the transformation manually to what I get from
>>>> cairo_fill_extents() but I don't seem to get it 100% right. It looks
>>>> like one row is missing at the bottom.
>>>
>>>> This is what I've tried:
>>>
>>>> cairo_t *cr;
>>>> cairo_surface_t *surface;
>>>> cairo_matrix_t cm;
>>>> int k;
>>>> double tx = 0, ty = 0;
>>>> double x1, y1, x2, y2;
>>>>
>>>> surface = cairo_image_surface_create(CAIRO_FORMAT_A8, 1000, 1000);
>>>> cr = cairo_create(surface);
>>>
>>>> for(k = 0; k < 2; k++) {
>>>>
>>>> cairo_new_path(cr);
>>>> cairo_move_to(cr, tx, ty);
>>>> cairo_set_font_size(cr, 100);
>>>> cairo_text_path(cr, "C");
>>>>
>>>> if(!k) {
>>>>
>>>> double width, height;
>>>>
>>>> cairo_fill_extents(cr, &x1, &y1, &x2, &y2);
>>>>
>>>> tx = -x1;
>>>> ty = -y1;
>>>>
>>>> width = fabs(x2 - x1);
>>>> height = fabs(y2 - y1);
>>>
>>>> width = ceil(width * 4.6);
>>>> height = ceil(height * 4.6);
>>>>
>>>> cairo_destroy(cr);
>>>> cairo_surface_destroy(surface);
>>>>
>>>> surface =
>>>> cairo_image_surface_create(CAIRO_FORMAT_A8, width, height);
>>>> cr = cairo_create(surface);
>>>>
>>>> cairo_matrix_init(&cm, 4.6, 0, 0, 4.6, 0, 0);
>>>> cairo_set_matrix(cr, &cm);
>>>> }
>>>> }
>>>>
>>>> cairo_fill(cr);
>>>
>>>> cairo_surface_flush(surface);
>>>> cairo_surface_write_to_png(surface, "tmp.png");
>>>>
>>>> cairo_destroy(cr);
>>>> cairo_surface_destroy(surface);
>>>
>>>> I'm attaching the resulting image. As you can see, there's at least
>>>> one row of the "C" shape missing at the bottom of the image.
>>>
>>>> Any ideas how to get this right? Is this some sort of floating
>>>> point inaccuracy and should I just add 1 to width/height to solve
>>>> this or am I doing something wrong here and there is a better way?
>>>
>>>
>>>
>>> --
>>> Best regards,
>>> Andreas Falkenhahn mailto:andreas at falkenhahn.com
>>>
>>>
>
>
>
>
> --
> Best regards,
> Andreas Falkenhahn mailto:andreas at falkenhahn.com
>
>
--
Best regards,
Andreas Falkenhahn mailto:andreas at falkenhahn.com
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