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Use circles for rendering light arc box-drawing characters.

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Ellipses can look weird when the font is narrow.

Closes #988
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L3MON4D3 2022-03-23 11:55:16 +01:00 • committed by Daniel Eklöf
parent bbf9dcc2a3
commit d6dab2f2ba
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CHANGELOG.md
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@ -60,6 +60,8 @@
(https://codeberg.org/dnkl/foot/issues/971).
* `\r` is no longer translated to `\n` when pasting clipboard data
(https://codeberg.org/dnkl/foot/issues/980).
* Use circles for rendering light arc box-drawing characters
(https://codeberg.org/dnkl/foot/issues/988).
### Deprecated

328
box-drawing.c
View file

@ -15,6 +15,8 @@
#include "util.h"
#include "xmalloc.h"
#define clamp(x, lower, upper) (min(upper, max(x, lower)))
enum thickness {
LIGHT,
HEAVY,
@ -1271,24 +1273,196 @@ draw_box_drawings_light_arc(struct buf *buf, char32_t wc)
const int thick = thickness(LIGHT) * supersample;
const bool thick_is_odd = (thick / supersample) % 2;
const bool height_is_odd = buf->height % 2;
const bool width_is_odd = buf->width % 2;
const int height_pixels = buf->height;
const int width_pixels = buf->width;
const int thick_pixels = thickness(LIGHT);
const double a = (width - thick) / 2;
const double b = (height - thick) / 2;
/*
* The general idea here is to connect the two incoming lines using a
* circle, which is extended to the box-edges with vertical/horizontal
* lines.
*
* The radius of the quartercircle should be as big as possible, with some
* restrictions: The radius should be the same for all of ╭ ╮ ╯ ╰ at a
* given box-size (which won't be the case if we choose the biggest
* possible radius for a given box, consider the following:)
*
*
* ▕ ███ ▏
* ▕a │ d▔▔
* ▕ │ x
* ▕ │
* ▕ │ ██
* ▕ ╰────██
* ▕ ██
* ▕
* ▕
* ▕b c
* ▔▔▔▔▔▔▔▔▔▔
* for ╰ it would be possible to center the circle on the upper right
* corner of d, but we have set it on x instead because ╯ can only use a
* 2px inner radius:
*
* ▕ ███ ▏
* ▔▔a │ d▏
* x │ ▏
* │ ▏
* ██ │ ▏
* ██───╯ ▏
* ██ ▏
* ▏
* ▏
* b c▏
* ▔▔▔▔▔▔▔▔▔▔
* As the incoming lines always exactly fill pixels, and are rounded down
* (via float->int), we can use this to get the radius of the inner
* (connecting the left/upper edge of the lines) quartercircle.
*/
int circle_inner_edge = (min(width_pixels, height_pixels) - thick_pixels) / 2;
const double a2 = a * a;
const double b2 = b * b;
/*
* We want to draw the quartercircle by filling small circles (with r =
* thickness/2.) whose centers are on its edge. This means to get the
* radius of the quartercircle, we add the exact half thickness to the
* radius of the inner circle.
*/
double c_r = circle_inner_edge + thick_pixels/2.;
const int num_samples = height * 16;
/*
* We need to draw short lines from the end of the quartercircle to the
* box-edges, store one endpoint (the other is the edge of the
* quartercircle) in these vars.
*/
int vert_to = 0,
hor_to = 0;
for (int i = 0; i < num_samples; i++) {
/* Coordinates of the circle-center. */
int c_x = 0,
c_y = 0;
/*
* For a given y there are up to two solutions for the circle-equation.
* Set to -1 for the left, and 1 for the right hemisphere.
*/
int circle_hemisphere = 0;
/*
* The quarter circle only has to be evaluated for a small range of
* y-values.
*/
int y_min = 0,
y_max = 0;
switch (wc) {
case U'╭': {
/*
* Don't use supersampled coordinates yet, we want to align actual
* pixels.
*
* pixel-coordinates of the lower edge of the right line and the
* right edge of the bottom line.
*/
int right_bottom_edge = (height_pixels + thick_pixels) / 2;
int bottom_right_edge = (width_pixels + thick_pixels) / 2;
/* find coordinates of circle-center. */
c_y = right_bottom_edge + circle_inner_edge;
c_x = bottom_right_edge + circle_inner_edge;
/* we want to render the left, not the right hemisphere of the circle. */
circle_hemisphere = -1;
/* don't evaluate beyond c_y, the vertical line is drawn there. */
y_min = 0;
y_max = c_y;
/*
* the vertical line should extend to the bottom of the box, the
* horizontal to the right.
*/
vert_to = height_pixels;
hor_to = width_pixels;
break;
}
case U'╮': {
int left_bottom_edge = (height_pixels + thick_pixels) / 2;
int bottom_left_edge = (width_pixels - thick_pixels) / 2;
c_y = left_bottom_edge + circle_inner_edge;
c_x = bottom_left_edge - circle_inner_edge;
circle_hemisphere = 1;
y_min = 0;
y_max = c_y;
vert_to = height_pixels;
hor_to = 0;
break;
}
case U'╰': {
int right_top_edge = (height_pixels - thick_pixels) / 2;
int top_right_edge = (width_pixels + thick_pixels) / 2;
c_y = right_top_edge - circle_inner_edge;
c_x = top_right_edge + circle_inner_edge;
circle_hemisphere = -1;
y_min = c_y;
y_max = height_pixels;
vert_to = 0;
hor_to = width_pixels;
break;
}
case U'╯': {
int left_top_edge = (height_pixels - thick_pixels) / 2;
int top_left_edge = (width_pixels - thick_pixels) / 2;
c_y = left_top_edge - circle_inner_edge;
c_x = top_left_edge - circle_inner_edge;
circle_hemisphere = 1;
y_min = c_y;
y_max = height_pixels;
vert_to = 0;
hor_to = 0;
break;
}
}
/* store for horizontal+vertical line. */
int c_x_pixels = c_x;
int c_y_pixels = c_y;
/* Bring coordinates from pixel-grid to supersampled grid. */
c_r *= supersample;
c_x *= supersample;
c_y *= supersample;
y_min *= supersample;
y_max *= supersample;
double c_r2 = c_r * c_r;
/*
* To prevent gaps in the circle, each pixel is sampled multiple times.
* As the quartercircle ends (vertically) in the middle of a pixel, an
* uneven number helps hit that exactly.
*/
for (double i = y_min*16; i <= y_max*16; i++) {
errno = 0;
feclearexcept(FE_ALL_EXCEPT);
double y = i / 16.;
double x = sqrt(a2 * (1. - y * y / b2));
double x = circle_hemisphere * sqrt(c_r2 - (y - c_y) * (y - c_y)) + c_x;
/* See math_error(7) */
if (errno != 0 ||
@ -1303,78 +1477,39 @@ draw_box_drawings_light_arc(struct buf *buf, char32_t wc)
if (col < 0)
continue;
int row_start = 0;
int row_end = 0;
int col_start = 0;
int col_end = 0;
/* rectangle big enough to fit entire circle with radius thick/2. */
int row1 = row - (thick/2+1);
int row2 = row + (thick/2+1);
int col1 = col - (thick/2+1);
int col2 = col + (thick/2+1);
/*
* At this point, row/col is only correct for ╯. For the other
* arcs, we need to mirror the arc around either the x-, y- or
* both axis.
*
* When doing so, we need to adjust for asymmetrical cell
* dimensions.
*
* The amazing box drawing art below represents the lower part
* of a cell, with the beginning of a vertical line in the
* middle. Each column represents one pixel.
*
*
* Even cell Odd cell
*
* │ │ │ │
* Even line │ ┆ ┆ │ │ ┆ ┆ │
* │ ┆ ┆ │ │ ┆ ┆ │
* └─┴─┴─┴─┘ └─┴─┴─┴─┴─┘
*
*
* │ │ │ │
* Odd line │ ┆ ┆ │ │ ┆ ┆ │
* │ ┆ ┆ │ │ ┆ ┆ │
* └─┴─┴─┴─┘ └─┴─┴─┴─┴─┘
*
* As can be seen(?), the resulting line is asymmetrical when
* *either* the cell is odd sized, *or* the line is odd
* sized. But not when both are.
*
* Hence the ‘thick % 2 ^ width % 2’ in the expressions below.
*/
switch (wc) {
case U'╭':
row_end = height - row - (thick_is_odd ^ height_is_odd);
row_start = row_end - thick;
col_end = width - col - (thick_is_odd ^ width_is_odd);
col_start = col_end - thick;
break;
case U'╮':
row_end = height - row - (thick_is_odd ^ height_is_odd);
row_start = row_end - thick;
col_start = col - ((thick_is_odd ^ width_is_odd) ? supersample / 2 : 0);
col_end = col_start + thick;
break;
case U'╰':
row_start = row - ((thick_is_odd ^ height_is_odd) ? supersample / 2 : 0);
row_end = row_start + thick;
col_end = width - col - (thick_is_odd ^ width_is_odd);
col_start = col_end - thick;
break;
case U'╯':
row_start = row - ((thick_is_odd ^ height_is_odd) ? supersample / 2 : 0);
row_end = row_start + thick;
col_start = col - ((thick_is_odd ^ width_is_odd) ? supersample / 2 : 0);
col_end = col_start + thick;
break;
}
int row_start = min(row1, row2),
row_end = max(row1, row2),
col_start = min(col1, col2),
col_end = max(col1, col2);
xassert(row_end > row_start);
xassert(col_end > col_start);
/*
* draw circle with radius thick/2 around x,y.
* this is accomplished by rejecting pixels where the distance from
* their center to x,y is greater than thick/2.
*/
for (int r = max(row_start, 0); r < max(min(row_end, height), 0); r++) {
double r_midpoint = r + 0.5;
for (int c = max(col_start, 0); c < max(min(col_end, width), 0); c++) {
double c_midpoint = c + 0.5;
/* vector from point on quartercircle to midpoint of the current pixel. */
double center_midpoint_x = c_midpoint - x;
double center_midpoint_y = r_midpoint - y;
/* distance from current point to circle-center. */
double dist = sqrt(center_midpoint_x * center_midpoint_x + center_midpoint_y * center_midpoint_y);
/* skip if midpoint of pixel is outside the circle. */
if (dist > thick/2.)
continue;
if (fmt == PIXMAN_a1) {
size_t idx = c / 8;
size_t bit_no = c % 8;
@ -1385,45 +1520,6 @@ draw_box_drawings_light_arc(struct buf *buf, char32_t wc)
}
}
/*
* Since a cell may not be completely symmetrical around its y-
* and x-axis, the mirroring done above may result in the last
* col/row of the arc not being filled in. This code ensures they
* are.
*/
if (wc == U'╭' || wc == U'╰') {
for (int y = 0; y < thick; y++) {
int row = (height - thick) / 2 + y - ((thick_is_odd ^ height_is_odd) ? supersample / 2 : 0);
for (int col = width - supersample; col < width; col++) {
if (row >= 0 && row < height && col >= 0) {
if (fmt == PIXMAN_a1) {
size_t ofs = col / 8;
size_t bit_no = col % 8;
set_a1_bit(data, row * stride + ofs, bit_no);
} else
data[row * stride + col] = 0xff;
}
}
}
}
if (wc == U'╭' || wc == U'╮') {
for (int x = 0; x < thick; x++) {
int col = (width - thick) / 2 + x - ((thick_is_odd ^ width_is_odd) ? supersample / 2 : 0);
for (int row = height - supersample; row < height; row++) {
if (row >= 0 && col >= 0 && col < width) {
if (fmt == PIXMAN_a1) {
size_t ofs = col / 8;
size_t bit_no = col % 8;
set_a1_bit(data, row * stride + ofs, bit_no);
} else
data[row * stride + col] = 0xff;
}
}
}
}
if (fmt == PIXMAN_a8) {
xassert(data != buf->data);
@ -1442,6 +1538,10 @@ draw_box_drawings_light_arc(struct buf *buf, char32_t wc)
free(data);
}
/* draw vertical/horizontal lines from quartercircle-edge to box-edge. */
vline(min(c_y_pixels, vert_to), max(c_y_pixels, vert_to), (width_pixels - thick_pixels) / 2, thick_pixels);
hline(min(c_x_pixels, hor_to), max(c_x_pixels, hor_to), (height_pixels - thick_pixels) / 2, thick_pixels);
}
static void NOINLINE