Escape characters in grep

Question

I'm a little confused about how many backslashes are needed to escape the alternation operator | in regular expressions for grep. This

echo abcdef | grep -e"def|zzz"

outputs nothing, because grep is not in extended regex mode. Escaping with one backslash works,

echo abcdef | grep -e"def\|zzz"

prints abcdef. More surprisingly, escaping with 2 backslashes also works,

echo abcdef | grep -e"def\\|zzz"

prints abcdef. Escaping with three backslashes fails,

echo abcdef | grep -e"def\\\|zzz"

prints nothing.

Does anyone have an explanation, especially for the 2-backslash case ?

Edit:

Using this simple argument-printing program,

void main(int argc, char** argv)
{
    for (int i = 0; i < argc; i++)
        printf("Arg %d: %s\n", i, argv[i]);
}

I investigated what my shell does with the command lines above :

-e"def|zzz" becomes -edef|zzz

-e"def\|zzz" becomes -edef\|zzz

-e"def\\|zzz" becomes -edef\\|zzz

-e"def\\\|zzz" becomes -edef\\\|zzz

So all double-quotes are removed and the backslashes and pipes are not altered by the shell. I suspect grep itself does something special with the literal string \\|.

Answer 1

The lowercase -e option is used to express multiple search operations. The alternation is implied:

$ echo abcdef | grep -e 'def' -e'zzz'
abcdef
$ echo abczzz | grep -e 'def' -e'zzz'
abczzz

Alternatively, you can use the upper -E option for extended regular expression notation:

$ echo abcdef | grep -E 'def|zzz'
abcdef

I believe this solves you problem directly (either using -e for alternation or -E for extended regex notation). Hope this helps :-)

FWIW, the issue with the backslashes is that | has special meaning to bash and needs to be escaped unless it is in single quotes. Here is a resource on quoting and escaping rules and the common pitfalls: https://web.archive.org/web/20230323230844/https://wiki.bash-hackers.org/syntax/quoting