How can I write a Perl one-liner for a Makefile so I can translate a literal \n to a newline?

Question

I need to escape \n so on output I really get newline or tab

$ perl -p -e 's/e/a/ge'

now I want each e to be substituted with \n

$ perl -p -e 's/e/\n/ge'

but even \n gives me an error.

this was a simplified example. In real script(makefile) I have

substitute := perl -p -e 's/@([^@]+)@/defined $$ENV{$$1} ? $$ENV{$$1} : $$1/ge'

and in target I have such a nice command

$(substitute) $< > $@

and if the input file for perl contains \n at output I will see it literally... I want to have real newline.

Answer 1

Remove the e modifier and the substitution will work fine:

perl -p -e 's/e/\n/g' file

From perldoc perlop:

e   Evaluate the right side as an expression.

UPD: If you want to preserve it, put the escape sequence in double quotes:

perl -p -e 's/e/"\n"/ge' file

Answer 2

You don't really need e in your regexp:

dpavlin@llin:~/private/home_html$ perl -p -e 's/e/\n/g'
foedoe
fo
do

Answer 3

To interpolate things that themselves need interpolation (like variables or escapes), use a double e and add quotes:

perl -p -e 's/@([^@]+)@/q!"! . ( defined $ENV{$1} ? $ENV{$1} : $1 ) . q!"!/gee'

That won't work if the substitution itself contains an unescaped " or $ or @; in that case, you'll need to do the backslash-escape handling separately.

Answer 4

echo xeieio | perl -pe 's/(.)/$k=$1;$k=~s#e#\n#;$k/ge'

outputs:

x
i
io

What it does... for each character, assign it to $k, run a simple s/e/\n/ on $k, then print $k. (Also, as always, there's more than one way to do it)