Mixing
recursive matching for string delimiter with regular expression
In verilog language, the statements are enclosed in a begin-end delimiter instead of bracket.
always@ (*) begin
if (condA) begin
a = c
end
else begin
b = d
end
end
I'd like to parse outermost begin-end with its statements to check coding rule in python. Using regular expression, I want results with regular expression like:
if (condA) begin
a = c
end
else begin
b = d
end
I found similar answer for bracket delimiter.
int funcA() {
if (condA) {
b = a
}
}
regular expression:
/({(?>[^{}]+|(?R))*})/g
However, I don't know how to modify atomic group ([^{}]) for "begin-end"?
/(begin(?>[??????]+|(?R))*end)/g
regex recursion
asked Jan 2 at 11:15
Jonghun yooJonghun yoo
61
add a comment |
In verilog language, the statements are enclosed in a begin-end delimiter instead of bracket.
always@ (*) begin
if (condA) begin
a = c
end
else begin
b = d
end
end
I'd like to parse outermost begin-end with its statements to check coding rule in python. Using regular expression, I want results with regular expression like:
if (condA) begin
a = c
end
else begin
b = d
end
I found similar answer for bracket delimiter.
int funcA() {
if (condA) {
b = a
}
}
regular expression:
/({(?>[^{}]+|(?R))*})/g
However, I don't know how to modify atomic group ([^{}]) for "begin-end"?
/(begin(?>[??????]+|(?R))*end)/g
regex recursion
asked Jan 2 at 11:15
Jonghun yooJonghun yoo
61
*is a greedy quantifier. It will find the longest matching sequence. Maybe I misunderstand something, but I think this may work:begin([sS]*)endCheck here
– Jónás Balázs
Jan 2 at 11:27
add a comment |
In verilog language, the statements are enclosed in a begin-end delimiter instead of bracket.
always@ (*) begin
if (condA) begin
a = c
end
else begin
b = d
end
end
I'd like to parse outermost begin-end with its statements to check coding rule in python. Using regular expression, I want results with regular expression like:
if (condA) begin
a = c
end
else begin
b = d
end
I found similar answer for bracket delimiter.
int funcA() {
if (condA) {
b = a
}
}
regular expression:
/({(?>[^{}]+|(?R))*})/g
However, I don't know how to modify atomic group ([^{}]) for "begin-end"?
/(begin(?>[??????]+|(?R))*end)/g
regex recursion
asked Jan 2 at 11:15
Jonghun yooJonghun yoo
61
In verilog language, the statements are enclosed in a begin-end delimiter instead of bracket.
always@ (*) begin
if (condA) begin
a = c
end
else begin
b = d
end
end
I'd like to parse outermost begin-end with its statements to check coding rule in python. Using regular expression, I want results with regular expression like:
if (condA) begin
a = c
end
else begin
b = d
end
I found similar answer for bracket delimiter.
int funcA() {
if (condA) {
b = a
}
}
regular expression:
/({(?>[^{}]+|(?R))*})/g
However, I don't know how to modify atomic group ([^{}]) for "begin-end"?
/(begin(?>[??????]+|(?R))*end)/g
regex recursion
regex recursion
asked Jan 2 at 11:15
Jonghun yooJonghun yoo
61
asked Jan 2 at 11:15
Jonghun yooJonghun yoo
61
asked Jan 2 at 11:15
Jonghun yooJonghun yoo
61
asked Jan 2 at 11:15
Jonghun yooJonghun yoo
61
asked Jan 2 at 11:15
Jonghun yooJonghun yoo
61
61
*is a greedy quantifier. It will find the longest matching sequence. Maybe I misunderstand something, but I think this may work:begin([sS]*)endCheck here
– Jónás Balázs
Jan 2 at 11:27
add a comment |
*is a greedy quantifier. It will find the longest matching sequence. Maybe I misunderstand something, but I think this may work:begin([sS]*)endCheck here
– Jónás Balázs
Jan 2 at 11:27
* is a greedy quantifier. It will find the longest matching sequence. Maybe I misunderstand something, but I think this may work: begin([sS]*)end Check here– Jónás Balázs
Jan 2 at 11:27
* is a greedy quantifier. It will find the longest matching sequence. Maybe I misunderstand something, but I think this may work: begin([sS]*)end Check here– Jónás Balázs
Jan 2 at 11:27
add a comment |
1 Answer
1
active
oldest
votes
The point of the [??????]+ part is to match any text that does not match a char that is equal or is the starting point of the delimiters.
So, in your case, you need to match any char other than a char that starts either begin or end substring:
/begin(?>(?!begin|end).|(?R))*end/gs
See the regex demo
The . here will match any char including line break chars due to the s modifier. Note that the actual implementation might need adjustments (e.g. in PHP, the g modifier should not be used as there are specific functions/features for that).
Also, since you recurse the whole pattern, you need no outer parentheses.
answered Jan 2 at 11:18
Wiktor StribiżewWiktor Stribiżew
325k16146226
Awesome! All done, thanks to your help.
– Jonghun yoo
Jan 2 at 11:37
Unrolling the pattern may yield better performance, but it becomes too cryptic and unwieldly, seebegin(?:[^be]*(?:(?:b(?!egin)|e(?!nd))[^be]*)*|(?R))*end.
– Wiktor Stribiżew
Jan 2 at 11:44
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
The point of the [??????]+ part is to match any text that does not match a char that is equal or is the starting point of the delimiters.
So, in your case, you need to match any char other than a char that starts either begin or end substring:
/begin(?>(?!begin|end).|(?R))*end/gs
See the regex demo
The . here will match any char including line break chars due to the s modifier. Note that the actual implementation might need adjustments (e.g. in PHP, the g modifier should not be used as there are specific functions/features for that).
Also, since you recurse the whole pattern, you need no outer parentheses.
answered Jan 2 at 11:18
Wiktor StribiżewWiktor Stribiżew
325k16146226
Awesome! All done, thanks to your help.
– Jonghun yoo
Jan 2 at 11:37
Unrolling the pattern may yield better performance, but it becomes too cryptic and unwieldly, seebegin(?:[^be]*(?:(?:b(?!egin)|e(?!nd))[^be]*)*|(?R))*end.
– Wiktor Stribiżew
Jan 2 at 11:44
add a comment |
The point of the [??????]+ part is to match any text that does not match a char that is equal or is the starting point of the delimiters.
So, in your case, you need to match any char other than a char that starts either begin or end substring:
/begin(?>(?!begin|end).|(?R))*end/gs
See the regex demo
The . here will match any char including line break chars due to the s modifier. Note that the actual implementation might need adjustments (e.g. in PHP, the g modifier should not be used as there are specific functions/features for that).
Also, since you recurse the whole pattern, you need no outer parentheses.
answered Jan 2 at 11:18
Wiktor StribiżewWiktor Stribiżew
325k16146226
Awesome! All done, thanks to your help.
– Jonghun yoo
Jan 2 at 11:37
Unrolling the pattern may yield better performance, but it becomes too cryptic and unwieldly, seebegin(?:[^be]*(?:(?:b(?!egin)|e(?!nd))[^be]*)*|(?R))*end.
– Wiktor Stribiżew
Jan 2 at 11:44
add a comment |
The point of the [??????]+ part is to match any text that does not match a char that is equal or is the starting point of the delimiters.
So, in your case, you need to match any char other than a char that starts either begin or end substring:
/begin(?>(?!begin|end).|(?R))*end/gs
See the regex demo
The . here will match any char including line break chars due to the s modifier. Note that the actual implementation might need adjustments (e.g. in PHP, the g modifier should not be used as there are specific functions/features for that).
Also, since you recurse the whole pattern, you need no outer parentheses.
answered Jan 2 at 11:18
Wiktor StribiżewWiktor Stribiżew
325k16146226
The point of the [??????]+ part is to match any text that does not match a char that is equal or is the starting point of the delimiters.
So, in your case, you need to match any char other than a char that starts either begin or end substring:
/begin(?>(?!begin|end).|(?R))*end/gs
See the regex demo
The . here will match any char including line break chars due to the s modifier. Note that the actual implementation might need adjustments (e.g. in PHP, the g modifier should not be used as there are specific functions/features for that).
Also, since you recurse the whole pattern, you need no outer parentheses.
answered Jan 2 at 11:18
Wiktor StribiżewWiktor Stribiżew
325k16146226
answered Jan 2 at 11:18
Wiktor StribiżewWiktor Stribiżew
325k16146226
answered Jan 2 at 11:18
Wiktor StribiżewWiktor Stribiżew
325k16146226
answered Jan 2 at 11:18
Wiktor StribiżewWiktor Stribiżew
325k16146226
325k16146226
Awesome! All done, thanks to your help.
– Jonghun yoo
Jan 2 at 11:37
Unrolling the pattern may yield better performance, but it becomes too cryptic and unwieldly, seebegin(?:[^be]*(?:(?:b(?!egin)|e(?!nd))[^be]*)*|(?R))*end.
– Wiktor Stribiżew
Jan 2 at 11:44
add a comment |
Awesome! All done, thanks to your help.
– Jonghun yoo
Jan 2 at 11:37
Unrolling the pattern may yield better performance, but it becomes too cryptic and unwieldly, seebegin(?:[^be]*(?:(?:b(?!egin)|e(?!nd))[^be]*)*|(?R))*end.
– Wiktor Stribiżew
Jan 2 at 11:44
Awesome! All done, thanks to your help.
– Jonghun yoo
Jan 2 at 11:37
Awesome! All done, thanks to your help.
– Jonghun yoo
Jan 2 at 11:37
Unrolling the pattern may yield better performance, but it becomes too cryptic and unwieldly, see
begin(?:[^be]*(?:(?:b(?!egin)|e(?!nd))[^be]*)*|(?R))*end.– Wiktor Stribiżew
Jan 2 at 11:44
Unrolling the pattern may yield better performance, but it becomes too cryptic and unwieldly, see
begin(?:[^be]*(?:(?:b(?!egin)|e(?!nd))[^be]*)*|(?R))*end.– Wiktor Stribiżew
Jan 2 at 11:44
add a comment |
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*is a greedy quantifier. It will find the longest matching sequence. Maybe I misunderstand something, but I think this may work:begin([sS]*)endCheck here– Jónás Balázs
Jan 2 at 11:27