Can't realize alias/substitution function for my .bashrc [duplicate] - linux

Question

This question already has answers here:
Bash script to receive and repass quoted parameters
(2 answers)
Closed 3 years ago.
I'm trying to colourize my find command so I've added this alias function to my .bashrc.
# liberate your find
function find
{
command find $# -exec ls --color=auto -d {} \;
}
But there is unexpected behavior using this code. It drops my quotes.
GNU bash, version 4.4.23(1)-release (x86_64-pc-linux-gnu)
Use my function:
find ./ -name '*.pl' -or -name '*.pm'
Result:
./lib/cover.pm
./lib/db.pm
Using the same find function but built-in:
command find ./ -name '*.pl' -or -name '*.pm'
Result:
./auth.pl
./index.pl
./title.pl
./lib/cover.pm
./lib/db.pm
./fs2db.pl
So the second variant didn't eat my quotes and works as it should.

Answer 1

For reproducing the problem I created all the files as shown in the longer Result in the question.
When I define the function(*) as
function find
{
command find $# -exec ls --color=auto -d {} \;
}
and execute
find ./ -name '*.pl' -or -name '*.pm'
I get an error message
find: paths must precede expression: fs2db.pl
Usage: find [-H] [-L] [-P] [-Olevel] [-D help|tree|search|stat|rates|opt|exec] [path...] [expression]
because *.pl gets expanded to auth.pl fs2db.pl index.pl title.pl by the shell.
I had to change the function to
function find
{
command find "$#" -exec ls --color=auto -d {} \;
}
to reproduce your problem. (Maybe this depends on the shell. I tested with bash 4.4.19(3)-release)
After set -x you can see what the shell does when executing your function:
$ find ./ -name '*.pl' -or -name '*.pm'
+ find ./ -name '*.pl' -or -name '*.pm'
+ command find ./ -name '*.pl' -or -name '*.pm' -exec ls --color=auto -d '{}' ';'
+ find ./ -name '*.pl' -or -name '*.pm' -exec ls --color=auto -d '{}' ';'
./lib/cover.pm
./lib/db.pm
The difference between executing your function and executing the find command directly is that your function appends an -exec action with the implicit -a (AND) operator. Without an explicit action, find prints all matching results.
You see a result of the operator precedence -a (AND) higher than -o (=-or, OR)
You can compare the output of these 3 commands
command find ./ -name '*.pl' -or -name '*.pm'
command find ./ -name '*.pl' -or -name '*.pm' -print
command find ./ \( -name '*.pl' -or -name '*.pm' \) -print
see http://man7.org/linux/man-pages/man1/find.1.html#NON-BUGS
You can call your function as
find ./ \( -name '*.pl' -or -name '*.pm' \)
to avoid the problem.
(*) This function definition is copied from the question.
You should use the portable POSIX style find() { ... } instead, unless there is a specific requirement for the Korn shell style function find { ... }.

Answer 2

As written, the -exec primary only applies to the code on the right of the -or operator. You need to parenthesize your arguments so that -exec applies to everything that matches. You also need to extract the path from the other arguments (which gets messy if you want to specify multiple paths, as your function would have to decide where to put the parentheses; distinguishing between paths and other expressions would amount to reimplementing a good chunk of find's parsing. I'll assume you are only passing a single path here).
find ()
{
path=$1
shift
command find "$path" \( "$#" \) -exec ls --color=auto -d {} \;
}
Alternatively, you can put the parentheses in the command line, with no change to your current definition.
find ./ \( -name '*.pl' -or -name '*.pm' \)
Your original function runs
find ./ -name '*.pl' -or -name '*.pm' -exec ls --color=auto -d {} \;
which is equivalent to
find ./ -name '*.pl' -or \( -name '*.pm' -exec ls --color=auto -d {} \; \)
with no implicit -print.