Avoid one character names in myers

This commit is contained in:
Armin Ronacher 2021-01-17 23:14:30 +01:00
parent e556888ce5
commit 0fb88ebe7d

View file

@ -68,58 +68,60 @@ where
New::Output: PartialEq<Old::Output>,
{
if old_end > old_current && new_end > new_current {
let n = old_end - old_current;
let m = new_end - new_current;
let l = (n + m) as isize;
let z = (2 * min(n, m) + 2) as usize;
let w = n as isize - m as isize;
let mut g = vec![0; z as usize];
let mut p = vec![0; z as usize];
for h in 0..=(l / 2 + l % 2) {
let old_span = old_end - old_current;
let new_span = new_end - new_current;
let total_span = (old_span + new_span) as isize;
let vec_size = (2 * min(old_span, new_span) + 2) as usize;
let w = old_span as isize - new_span as isize;
let mut vec_down = vec![0; vec_size as usize];
let mut vec_up = vec![0; vec_size as usize];
for i in 0..=(total_span / 2 + total_span % 2) {
for &inverse in &[true, false][..] {
let (dollar_c, dollar_d) = if inverse {
(&mut g, &mut p)
(&mut vec_down, &mut vec_up)
} else {
(&mut p, &mut g)
(&mut vec_up, &mut vec_down)
};
let (k0, k1) = {
let (m, n) = (m as isize, n as isize);
(-(h - 2 * max(0, h - m)), h - 2 * max(0, h - n) + 1)
let (m, n) = (new_span as isize, old_span as isize);
(-(i - 2 * max(0, i - m)), i - 2 * max(0, i - n) + 1)
};
for k in (k0..k1).step_by(2) {
let mut a: usize = if k == -h
|| k != h && dollar_c[modulo(k - 1, z)] < dollar_c[modulo(k + 1, z)]
let mut a: usize = if k == -i
|| k != i
&& dollar_c[modulo(k - 1, vec_size)] < dollar_c[modulo(k + 1, vec_size)]
{
dollar_c[modulo(k + 1, z)]
dollar_c[modulo(k + 1, vec_size)]
} else {
dollar_c[modulo(k - 1, z)] + 1
dollar_c[modulo(k - 1, vec_size)] + 1
};
let mut b = (a as isize - k) as usize;
let (s, t) = (a, b);
while a < n && b < m && {
while a < old_span && b < new_span && {
let (e_i, f_i) = if inverse {
(a, b)
} else {
(n - a - 1, m - b - 1)
(old_span - a - 1, new_span - b - 1)
};
new[new_current + f_i] == old[old_current + e_i]
} {
a += 1;
b += 1;
}
dollar_c[modulo(k, z)] = a;
let bound = if inverse { h - 1 } else { h };
if (l % 2 == 1) == inverse
dollar_c[modulo(k, vec_size)] = a;
let bound = if inverse { i - 1 } else { i };
if (total_span % 2 == 1) == inverse
&& w - k >= -bound
&& w - k <= bound
&& dollar_c[modulo(k, z)] + dollar_d[modulo(w - k, z)] >= n
&& dollar_c[modulo(k, vec_size)] + dollar_d[modulo(w - k, vec_size)]
>= old_span
{
let (x, y, u, v) = if inverse {
(s, t, a, b)
} else {
(n - a, m - b, n - s, m - t)
(old_span - a, new_span - b, old_span - s, new_span - t)
};
if h + bound > 1 || (x != u && y != v) {
if i + bound > 1 || (x != u && y != v) {
diff_offsets(
diff,
old,
@ -142,13 +144,21 @@ where
new_end,
)?;
return Ok(());
} else if m > n {
diff.equal(old_current, new_current, n)?;
diff.insert(old_current + n, new_current + n, m - n)?;
} else if new_span > old_span {
diff.equal(old_current, new_current, old_span)?;
diff.insert(
old_current + old_span,
new_current + old_span,
new_span - old_span,
)?;
return Ok(());
} else if m < n {
diff.equal(old_current, new_current, m)?;
diff.delete(old_current + m, n - m, new_current + m)?;
} else if new_span < old_span {
diff.equal(old_current, new_current, new_span)?;
diff.delete(
old_current + new_span,
old_span - new_span,
new_current + new_span,
)?;
return Ok(());
} else {
return Ok(());