在进行嵌套分支重置时，我已经尽可能地猜测它是如何工作的。我在互联网上搜索了嵌套信息，但找不到明确的确认信息。
大多数情况下，关注的是在内部嵌套时立即发生的顺序 以下样本是我最好的猜测，如果有人能够确认其正确或引导我朝着正确的方向进行，那么它将会被认可。
示例正则表达式：
(a)(?|x(y)z(?|(u)(u)(u)(u)(u)(u)|(e)(e)(e)|(c))(K)|(p(q(?|(M)(M)(M)(M)(?|(T)(T)(T)|(D)(D))(R)(R)|(B)(B)(B)|(v)))r)(o)(i)|(t)s(w))(Z)
Number Sequenced regex：
1 ( a )
(?|
x
2 ( y )
z
(?|
3 ( u )
4 ( u )
5 ( u )
6 ( u )
7 ( u )
8 ( u )
|
3 ( e )
4 ( e )
5 ( e )
|
3 ( c )
)
9 ( K )
|
2 (
p
3 (
q
(?|
4 ( M )
5 ( M )
6 ( M )
7 ( M )
(?|
8 ( T )
9 ( T )
10 ( T )
|
8 ( D )
9 ( D )
)
11 ( R )
12 ( R )
|
4 ( B )
5 ( B )
6 ( B )
|
4 ( v )
)
3 )
r
2 )
13 ( o )
14 ( i )
|
2 ( t )
s
3 ( w )
)
15 ( Z )
Perl测试用例：
# (a)(?|x(y)z(?|(u)(u)(u)(u)(u)(u)|(e)(e)(e)|(c))(K)|(p(q(?|(M)(M)(M)(M)(?|(T)(T)(T)|(D)(D))(R)(R)|(B)(B)(B)|(v)))r)(o)(i)|(t)s(w))(Z)
( a ) # (1)
(?|
x
( y ) # (2)
z
(?|
( u ) # (3)
( u ) # (4)
( u ) # (5)
( u ) # (6)
( u ) # (7)
( u ) # (8)
|
( e ) # (3)
( e ) # (4)
( e ) # (5)
|
( c ) # (3)
)
( K ) # (9)
|
( # (2 start)
p
( # (3 start)
q
(?|
( M ) # (4)
( M ) # (5)
( M ) # (6)
( M ) # (7)
(?|
( T ) # (8)
( T ) # (9)
( T ) # (10)
|
( D ) # (8)
( D ) # (9)
)
( R ) # (11)
( R ) # (12)
|
( B ) # (4)
( B ) # (5)
( B ) # (6)
|
( v ) # (4)
)
) # (3 end)
r
) # (2 end)
( o ) # (13)
( i ) # (14)
|
( t ) # (2)
s
( w ) # (3)
)
( Z ) # (15)
Perl引擎结果：
输入
axyzuuuuuuKZ
axyzeeeKZ
axyzcKZ
apqMMMMTTTRRroiZ
apqMMMMDDRRroiZ
apqBBBroiZ
apqvroiZ
atswZ
输出
** Grp 0 - ( pos 0 , len 12 )
axyzuuuuuuKZ
** Grp 1 - ( pos 0 , len 1 )
a
** Grp 2 - ( pos 2 , len 1 )
y
** Grp 3 - ( pos 4 , len 1 )
u
** Grp 4 - ( pos 5 , len 1 )
u
** Grp 5 - ( pos 6 , len 1 )
u
** Grp 6 - ( pos 7 , len 1 )
u
** Grp 7 - ( pos 8 , len 1 )
u
** Grp 8 - ( pos 9 , len 1 )
u
** Grp 9 - ( pos 10 , len 1 )
K
** Grp 10 - NULL
** Grp 11 - NULL
** Grp 12 - NULL
** Grp 13 - NULL
** Grp 14 - NULL
** Grp 15 - ( pos 11 , len 1 )
Z
-----------------------
** Grp 0 - ( pos 14 , len 9 )
axyzeeeKZ
** Grp 1 - ( pos 14 , len 1 )
a
** Grp 2 - ( pos 16 , len 1 )
y
** Grp 3 - ( pos 18 , len 1 )
e
** Grp 4 - ( pos 19 , len 1 )
e
** Grp 5 - ( pos 20 , len 1 )
e
** Grp 6 - NULL
** Grp 7 - NULL
** Grp 8 - NULL
** Grp 9 - ( pos 21 , len 1 )
K
** Grp 10 - NULL
** Grp 11 - NULL
** Grp 12 - NULL
** Grp 13 - NULL
** Grp 14 - NULL
** Grp 15 - ( pos 22 , len 1 )
Z
-----------------------
** Grp 0 - ( pos 25 , len 7 )
axyzcKZ
** Grp 1 - ( pos 25 , len 1 )
a
** Grp 2 - ( pos 27 , len 1 )
y
** Grp 3 - ( pos 29 , len 1 )
c
** Grp 4 - NULL
** Grp 5 - NULL
** Grp 6 - NULL
** Grp 7 - NULL
** Grp 8 - NULL
** Grp 9 - ( pos 30 , len 1 )
K
** Grp 10 - NULL
** Grp 11 - NULL
** Grp 12 - NULL
** Grp 13 - NULL
** Grp 14 - NULL
** Grp 15 - ( pos 31 , len 1 )
Z
-----------------------
** Grp 0 - ( pos 34 , len 16 )
apqMMMMTTTRRroiZ
** Grp 1 - ( pos 34 , len 1 )
a
** Grp 2 - ( pos 35 , len 12 )
pqMMMMTTTRRr
** Grp 3 - ( pos 36 , len 10 )
qMMMMTTTRR
** Grp 4 - ( pos 37 , len 1 )
M
** Grp 5 - ( pos 38 , len 1 )
M
** Grp 6 - ( pos 39 , len 1 )
M
** Grp 7 - ( pos 40 , len 1 )
M
** Grp 8 - ( pos 41 , len 1 )
T
** Grp 9 - ( pos 42 , len 1 )
T
** Grp 10 - ( pos 43 , len 1 )
T
** Grp 11 - ( pos 44 , len 1 )
R
** Grp 12 - ( pos 45 , len 1 )
R
** Grp 13 - ( pos 47 , len 1 )
o
** Grp 14 - ( pos 48 , len 1 )
i
** Grp 15 - ( pos 49 , len 1 )
Z
-----------------------
** Grp 0 - ( pos 52 , len 15 )
apqMMMMDDRRroiZ
** Grp 1 - ( pos 52 , len 1 )
a
** Grp 2 - ( pos 53 , len 11 )
pqMMMMDDRRr
** Grp 3 - ( pos 54 , len 9 )
qMMMMDDRR
** Grp 4 - ( pos 55 , len 1 )
M
** Grp 5 - ( pos 56 , len 1 )
M
** Grp 6 - ( pos 57 , len 1 )
M
** Grp 7 - ( pos 58 , len 1 )
M
** Grp 8 - ( pos 59 , len 1 )
D
** Grp 9 - ( pos 60 , len 1 )
D
** Grp 10 - NULL
** Grp 11 - ( pos 61 , len 1 )
R
** Grp 12 - ( pos 62 , len 1 )
R
** Grp 13 - ( pos 64 , len 1 )
o
** Grp 14 - ( pos 65 , len 1 )
i
** Grp 15 - ( pos 66 , len 1 )
Z
-----------------------
** Grp 0 - ( pos 69 , len 10 )
apqBBBroiZ
** Grp 1 - ( pos 69 , len 1 )
a
** Grp 2 - ( pos 70 , len 6 )
pqBBBr
** Grp 3 - ( pos 71 , len 4 )
qBBB
** Grp 4 - ( pos 72 , len 1 )
B
** Grp 5 - ( pos 73 , len 1 )
B
** Grp 6 - ( pos 74 , len 1 )
B
** Grp 7 - NULL
** Grp 8 - NULL
** Grp 9 - NULL
** Grp 10 - NULL
** Grp 11 - NULL
** Grp 12 - NULL
** Grp 13 - ( pos 76 , len 1 )
o
** Grp 14 - ( pos 77 , len 1 )
i
** Grp 15 - ( pos 78 , len 1 )
Z
-----------------------
** Grp 0 - ( pos 81 , len 8 )
apqvroiZ
** Grp 1 - ( pos 81 , len 1 )
a
** Grp 2 - ( pos 82 , len 4 )
pqvr
** Grp 3 - ( pos 83 , len 2 )
qv
** Grp 4 - ( pos 84 , len 1 )
v
** Grp 5 - NULL
** Grp 6 - NULL
** Grp 7 - NULL
** Grp 8 - NULL
** Grp 9 - NULL
** Grp 10 - NULL
** Grp 11 - NULL
** Grp 12 - NULL
** Grp 13 - ( pos 86 , len 1 )
o
** Grp 14 - ( pos 87 , len 1 )
i
** Grp 15 - ( pos 88 , len 1 )
Z
-----------------------
** Grp 0 - ( pos 91 , len 5 )
atswZ
** Grp 1 - ( pos 91 , len 1 )
a
** Grp 2 - ( pos 92 , len 1 )
t
** Grp 3 - ( pos 94 , len 1 )
w
** Grp 4 - NULL
** Grp 5 - NULL
** Grp 6 - NULL
** Grp 7 - NULL
** Grp 8 - NULL
** Grp 9 - NULL
** Grp 10 - NULL
** Grp 11 - NULL
** Grp 12 - NULL
** Grp 13 - NULL
** Grp 14 - NULL
** Grp 15 - ( pos 95 , len 1 )
Z
答案 0 :(得分：2)
似乎正确。由于分支重置中的捕获组数量等于其任何分支中捕获组的最大数量。
以下是perlre的引用：
每个分支内的编号将正常，并且此构造后面的任何组都将被编号，就好像构造只包含一个分支，即其中捕获组最多的分支。