/(.+)n1/ works but /(.*)n1/ doesn't when they should both work

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I was playing around with sed after answering another question and I noticed that .+ and .* are not giving the same result when they both match multiple characters in a context address.

The following command¹:

sed -E '$!N;/(.+)n1/!P;D' <<IN
one
one_more
two
two_more
IN

prints

one_more
two_more

OK, that's the expected output.

Changing the regex from .+ to .* (i.e. from one or more characters to zero or more characters) should give the same result but it does not:

sed -E '$!N;/(.*)n1/!P;D' <<IN
one
one_more
two
two_more
IN

prints just one line

two_more

What's going on here ?

^{1: I'm using ERE for simplicity/readability, the same happens when using BRE}

edited 1 hour ago

asked 3 hours ago

don_crissti

48.2k15127157

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up vote
4
down vote

favorite

sed -E '$!N;/(.+)n1/!P;D' <<IN
one
one_more
two
two_more
IN

prints

one_more
two_more

OK, that's the expected output.

Changing the regex from .+ to .* (i.e. from one or more characters to zero or more characters) should give the same result but it does not:

sed -E '$!N;/(.*)n1/!P;D' <<IN
one
one_more
two
two_more
IN

prints just one line

two_more

What's going on here ?

^{1: I'm using ERE for simplicity/readability, the same happens when using BRE}

edited 1 hour ago

asked 3 hours ago

don_crissti

48.2k15127157

add a commentÂ |Â

up vote
4
down vote

favorite

sed -E '$!N;/(.+)n1/!P;D' <<IN
one
one_more
two
two_more
IN

prints

one_more
two_more

OK, that's the expected output.

Changing the regex from .+ to .* (i.e. from one or more characters to zero or more characters) should give the same result but it does not:

sed -E '$!N;/(.*)n1/!P;D' <<IN
one
one_more
two
two_more
IN

prints just one line

two_more

What's going on here ?

^{1: I'm using ERE for simplicity/readability, the same happens when using BRE}

edited 1 hour ago

asked 3 hours ago

don_crissti

48.2k15127157

sed -E '$!N;/(.+)n1/!P;D' <<IN
one
one_more
two
two_more
IN

prints

one_more
two_more

OK, that's the expected output.

Changing the regex from .+ to .* (i.e. from one or more characters to zero or more characters) should give the same result but it does not:

sed -E '$!N;/(.*)n1/!P;D' <<IN
one
one_more
two
two_more
IN

prints just one line

two_more

What's going on here ?

^{1: I'm using ERE for simplicity/readability, the same happens when using BRE}

sed regular-expression

edited 1 hour ago

asked 3 hours ago

don_crissti

48.2k15127157

edited 1 hour ago

asked 3 hours ago

don_crissti

48.2k15127157

edited 1 hour ago

asked 3 hours ago

don_crissti

48.2k15127157

asked 3 hours ago

don_crissti

48.2k15127157

asked 3 hours ago

don_crissti

48.2k15127157

add a commentÂ |Â

1 Answer
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That happens because /(.*)n1/ is also matching a simple newline (n: empty string, followed by newline, followed by the same empty string from the beginning).

So it will also match the string one_morentwo from your example.

To avoid that, you'll have to anchor your regexps, e.g. sed -E '$!N;/^(.+)n1/!P;D' or sed -E '$!N;/^(.*)n1/!P;D'.

edited 2 hours ago

don_crissti

48.2k15127157

answered 2 hours ago

mosvy

3,115118

But wouldn't that contradict greediness?
â€“Â RudiC
2 hours ago

No. It's simply that "one_morentwo" will match /(.*)n1/, whether greedy or not. To understand what happens in that sed script, add a debugging command after the N eg: sed -E '$!N; h;s/(.; /(.*)n1/!P; D' ...
â€“Â mosvy
2 hours ago

@RudiC - greedy means the regex engine attempts to match as much as possible; in this particular case ((.*)n1) nothing (the empty string) followed by newline and again by nothing matches and that is sufficient.
â€“Â don_crissti
1 hour ago

@RudiC one_morentwo doesn't match /(.+)n1/ at all (since there's no non-empty string that appears on both sides of the NL), but it does match /(.*)n1/. Then again, one_moreneno would match /(.+)n1/ too, so perhaps the anchors are a good idea.
â€“Â ilkkachu
1 hour ago

add a commentÂ |Â

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1 Answer
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active

oldest

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up vote
5
down vote

accepted

That happens because /(.*)n1/ is also matching a simple newline (n: empty string, followed by newline, followed by the same empty string from the beginning).

So it will also match the string one_morentwo from your example.

To avoid that, you'll have to anchor your regexps, e.g. sed -E '$!N;/^(.+)n1/!P;D' or sed -E '$!N;/^(.*)n1/!P;D'.

edited 2 hours ago

don_crissti

48.2k15127157

answered 2 hours ago

mosvy

3,115118

But wouldn't that contradict greediness?
â€“Â RudiC
2 hours ago

No. It's simply that "one_morentwo" will match /(.*)n1/, whether greedy or not. To understand what happens in that sed script, add a debugging command after the N eg: sed -E '$!N; h;s/(.; /(.*)n1/!P; D' ...
â€“Â mosvy
2 hours ago

@RudiC - greedy means the regex engine attempts to match as much as possible; in this particular case ((.*)n1) nothing (the empty string) followed by newline and again by nothing matches and that is sufficient.
â€“Â don_crissti
1 hour ago

@RudiC one_morentwo doesn't match /(.+)n1/ at all (since there's no non-empty string that appears on both sides of the NL), but it does match /(.*)n1/. Then again, one_moreneno would match /(.+)n1/ too, so perhaps the anchors are a good idea.
â€“Â ilkkachu
1 hour ago

add a commentÂ |Â

up vote
5
down vote

accepted

That happens because /(.*)n1/ is also matching a simple newline (n: empty string, followed by newline, followed by the same empty string from the beginning).

So it will also match the string one_morentwo from your example.

To avoid that, you'll have to anchor your regexps, e.g. sed -E '$!N;/^(.+)n1/!P;D' or sed -E '$!N;/^(.*)n1/!P;D'.

edited 2 hours ago

don_crissti

48.2k15127157

answered 2 hours ago

mosvy

3,115118

But wouldn't that contradict greediness?
â€“Â RudiC
2 hours ago

No. It's simply that "one_morentwo" will match /(.*)n1/, whether greedy or not. To understand what happens in that sed script, add a debugging command after the N eg: sed -E '$!N; h;s/(.; /(.*)n1/!P; D' ...
â€“Â mosvy
2 hours ago

@RudiC - greedy means the regex engine attempts to match as much as possible; in this particular case ((.*)n1) nothing (the empty string) followed by newline and again by nothing matches and that is sufficient.
â€“Â don_crissti
1 hour ago

@RudiC one_morentwo doesn't match /(.+)n1/ at all (since there's no non-empty string that appears on both sides of the NL), but it does match /(.*)n1/. Then again, one_moreneno would match /(.+)n1/ too, so perhaps the anchors are a good idea.
â€“Â ilkkachu
1 hour ago

add a commentÂ |Â

up vote
5
down vote

accepted

That happens because /(.*)n1/ is also matching a simple newline (n: empty string, followed by newline, followed by the same empty string from the beginning).

So it will also match the string one_morentwo from your example.

To avoid that, you'll have to anchor your regexps, e.g. sed -E '$!N;/^(.+)n1/!P;D' or sed -E '$!N;/^(.*)n1/!P;D'.

edited 2 hours ago

don_crissti

48.2k15127157

answered 2 hours ago

mosvy

3,115118

That happens because /(.*)n1/ is also matching a simple newline (n: empty string, followed by newline, followed by the same empty string from the beginning).

So it will also match the string one_morentwo from your example.

To avoid that, you'll have to anchor your regexps, e.g. sed -E '$!N;/^(.+)n1/!P;D' or sed -E '$!N;/^(.*)n1/!P;D'.

edited 2 hours ago

don_crissti

48.2k15127157

answered 2 hours ago

mosvy

3,115118

edited 2 hours ago

don_crissti

48.2k15127157

edited 2 hours ago

don_crissti

48.2k15127157

edited 2 hours ago

don_crissti

48.2k15127157

answered 2 hours ago

mosvy

3,115118

answered 2 hours ago

mosvy

3,115118

answered 2 hours ago

mosvy

3,115118

But wouldn't that contradict greediness?
â€“Â RudiC
2 hours ago

No. It's simply that "one_morentwo" will match /(.*)n1/, whether greedy or not. To understand what happens in that sed script, add a debugging command after the N eg: sed -E '$!N; h;s/(.; /(.*)n1/!P; D' ...
â€“Â mosvy
2 hours ago

@RudiC - greedy means the regex engine attempts to match as much as possible; in this particular case ((.*)n1) nothing (the empty string) followed by newline and again by nothing matches and that is sufficient.
â€“Â don_crissti
1 hour ago

@RudiC one_morentwo doesn't match /(.+)n1/ at all (since there's no non-empty string that appears on both sides of the NL), but it does match /(.*)n1/. Then again, one_moreneno would match /(.+)n1/ too, so perhaps the anchors are a good idea.
â€“Â ilkkachu
1 hour ago

add a commentÂ |Â

But wouldn't that contradict greediness?
â€“Â RudiC
2 hours ago

No. It's simply that "one_morentwo" will match /(.*)n1/, whether greedy or not. To understand what happens in that sed script, add a debugging command after the N eg: sed -E '$!N; h;s/(.; /(.*)n1/!P; D' ...
â€“Â mosvy
2 hours ago

@RudiC - greedy means the regex engine attempts to match as much as possible; in this particular case ((.*)n1) nothing (the empty string) followed by newline and again by nothing matches and that is sufficient.
â€“Â don_crissti
1 hour ago

@RudiC one_morentwo doesn't match /(.+)n1/ at all (since there's no non-empty string that appears on both sides of the NL), but it does match /(.*)n1/. Then again, one_moreneno would match /(.+)n1/ too, so perhaps the anchors are a good idea.
â€“Â ilkkachu
1 hour ago

But wouldn't that contradict greediness?
â€“Â RudiC
2 hours ago

No. It's simply that "one_morentwo" will match /(.*)n1/, whether greedy or not. To understand what happens in that sed script, add a debugging command after the N eg: sed -E '$!N; h;s/(.; /(.*)n1/!P; D' ...
â€“Â mosvy
2 hours ago

@RudiC - greedy means the regex engine attempts to match as much as possible; in this particular case ((.*)n1) nothing (the empty string) followed by newline and again by nothing matches and that is sufficient.
â€“Â don_crissti
1 hour ago

@RudiC one_morentwo doesn't match /(.+)n1/ at all (since there's no non-empty string that appears on both sides of the NL), but it does match /(.*)n1/. Then again, one_moreneno would match /(.+)n1/ too, so perhaps the anchors are a good idea.
â€“Â ilkkachu
1 hour ago

add a commentÂ |Â

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