Why is the variable getting replaced with previous command in perl?

Question

I'm trying to replace only the first occurrence of a pattern in a file using a perl one liner.

>touch tmp
>perl -p -i.bak -e '++$seen if( !$seen && s/alpha/beta/);' tmp

After this I press Enter. Below message appears on the screen.

perl -p -i.bak -e '++$seen if( tmpseen && s/alpha/beta/);' tmp

Note that !$seen got replaced with tmpseen.

Why is this happening?

Thanks.

This is due to csh. !$ is expanding to the last argument of the previous command. I'm assuming that. — kjprice
– kjprice, Commented Mar 27, 2013 at 16:18
@kjprice indeed. that seems to be the case. what can I do now? — user13107
– user13107, Commented Mar 27, 2013 at 16:21
@user13107 I don't use csh, but I'm guessing if( \!$seen will work. — kjprice
– kjprice, Commented Mar 27, 2013 at 16:25

Jozef Legény · Accepted Answer · 2013-03-27 16:27:02Z

4

As kjprice mentioned !$ gets expanded by the shell. One possible solution is to use the operator 'not' instead of the operator !.

perl -p -i.bak -e '++$seen if( (not $seen) && s/alpha/beta/);' tmp

Parentheses are there because of the lower precedence of the operator 'not'

answered Mar 27, 2013 at 16:27

Jozef Legény

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Comments

kjprice · Accepted Answer · 2013-03-27 16:28:06Z

3

You can escape any csh special characters with \ :

 perl -p -i.bak -e '++$seen if( \!$seen && s/alpha/beta/);' tmp

answered Mar 27, 2013 at 16:28

kjprice

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