Bash – Order of Executables Started in Shell

bashcommandshellshell-builtin

If I execute the test command in bash, test(evaluates conditional expression) built-in utility is started:

$ type test
test is a shell builtin
$ type -a test
test is a shell builtin
test is /usr/local/bin/test
test is /usr/bin/test
$

However, as seen in output of type -a test above, there is another test in /usr/local/bin directory and yet another one in /usr/bin directory. How are executables ordered, i.e. are the built-in commands always preferred and then the rest of the commands depend on the directory order in $PATH variable? In addition, is it possible to change the order of the executables started, e.g. if I type in test, then /usr/bin/test is started instead of bash-builtin test?

Best Answer

Highest priority is bash alias, then special builtins (only in POSIX mode), then functions, then builtins, then a search in $PATH.

To execute a builtin, use builtin test.
To execute an external application, use an explicit path: /bin/test.
To ignore functions and aliases, use command test.
To bypass just alias, use \test or any other kind of expansion.

It's possible to disable/enable a builtin with enable test.

(Updated according to comments below)
(Fixed incorrect admin edit that bash has disable builtin - in fact, there is only enable)

Related Question