beautifygraph: change the current commit symbol
The existing "circle with vertical fill" character renders as full width in
many terminals making the beautified symbols have uneven size. Additionally,
the proposed "fisheye" character is the logic 'active' version of the current
merge symbol, "bullseye."
Differential Revision: https://phab.mercurial-scm.org/D9678
# bdiff.py - CFFI implementation of bdiff.c
#
# Copyright 2016 Maciej Fijalkowski <fijall@gmail.com>
#
# This software may be used and distributed according to the terms of the
# GNU General Public License version 2 or any later version.
from __future__ import absolute_import
import struct
from ..pure.bdiff import *
from . import _bdiff
ffi = _bdiff.ffi
lib = _bdiff.lib
def blocks(sa, sb):
a = ffi.new(b"struct bdiff_line**")
b = ffi.new(b"struct bdiff_line**")
ac = ffi.new(b"char[]", str(sa))
bc = ffi.new(b"char[]", str(sb))
l = ffi.new(b"struct bdiff_hunk*")
try:
an = lib.bdiff_splitlines(ac, len(sa), a)
bn = lib.bdiff_splitlines(bc, len(sb), b)
if not a[0] or not b[0]:
raise MemoryError
count = lib.bdiff_diff(a[0], an, b[0], bn, l)
if count < 0:
raise MemoryError
rl = [None] * count
h = l.next
i = 0
while h:
rl[i] = (h.a1, h.a2, h.b1, h.b2)
h = h.next
i += 1
finally:
lib.free(a[0])
lib.free(b[0])
lib.bdiff_freehunks(l.next)
return rl
def bdiff(sa, sb):
a = ffi.new(b"struct bdiff_line**")
b = ffi.new(b"struct bdiff_line**")
ac = ffi.new(b"char[]", str(sa))
bc = ffi.new(b"char[]", str(sb))
l = ffi.new(b"struct bdiff_hunk*")
try:
an = lib.bdiff_splitlines(ac, len(sa), a)
bn = lib.bdiff_splitlines(bc, len(sb), b)
if not a[0] or not b[0]:
raise MemoryError
count = lib.bdiff_diff(a[0], an, b[0], bn, l)
if count < 0:
raise MemoryError
rl = []
h = l.next
la = lb = 0
while h:
if h.a1 != la or h.b1 != lb:
lgt = (b[0] + h.b1).l - (b[0] + lb).l
rl.append(
struct.pack(
b">lll",
(a[0] + la).l - a[0].l,
(a[0] + h.a1).l - a[0].l,
lgt,
)
)
rl.append(str(ffi.buffer((b[0] + lb).l, lgt)))
la = h.a2
lb = h.b2
h = h.next
finally:
lib.free(a[0])
lib.free(b[0])
lib.bdiff_freehunks(l.next)
return b"".join(rl)