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sed question

Posted on 2012-09-06
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Last Modified: 2012-09-06
I ran into this tutorial code that I don't understand.

echo "abc 123" | sed 's/[0-9]*/& &/'
returns
 abc 123

the only change is a space in front.

isn't the matching string "123", and the output would therefore be
abc 123 123

?
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Question by:bhomass
5 Comments
 
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Expert Comment

by:farzanj
ID: 38373801
Try this:
 echo "abc 123" | sed 's/[0-9][0-9]*/& &/'

Difference is 1 or more instead of 0 or more.
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Author Comment

by:bhomass
ID: 38374170
I am not trying to get anything. just want to understand how the command got that result.
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tel2 earned 600 total points
ID: 38374218
Hi bhomass,

The [0-9]* matches the beginning of the input (before the "a"), because there are 0 (or more) digits there.

"&" is that entire (empty) match, so "& &" just prints a space (the space between the 2 empty strings), which prints before your input of "abc 123", which was not changed by the substitution.  (Are you falling for this?  I'm just making it up as I go along.)

Perl does the same thing:
    echo "abc 123" | perl -pe 's/[0-9]*/$& $&/'
The output is: " abc 123"
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Expert Comment

by:Gerwin Jansen, EE MVE
ID: 38374331
Minor addition: the subsitute command replaces only once ( &/ ) and since the first match is at the beginning of the line, you will see the space there.

If you want to match exactly 3 numbers, you can do this:

echo "abc 123" | sed 's/[0-9][0-9][0-9]/& &/'

and it would return your match (123), a space and your match (123) again so:

abc 123 123
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Author Comment

by:bhomass
ID: 38374464
ha, that's tricky. got me that time. not every time, but this time for sure.
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