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Asked: 2024-05-03 13:39:28 +0800 CST

`command echo --help` invokes bash internal echo function

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According to help command, the help of bash, command can be used to trigger the external program x, if x exists as a builtin function too:

command: command [-pVv] command [arg ...]
    Execute a simple command or display information about commands.
    
    Runs COMMAND with ARGS suppressing  shell function lookup, or display
    information about the specified COMMANDs.  Can be used to invoke commands
    on disk when a function with the same name exists.

When testing with time, this works as expected and uses the external program /bin/time, but when I try:

command echo --help    
# the output is 
--help

which is the expected behavior for the builtin echo.

Using xubuntu 22.04, bash 5.1.16 /bin/echo --help works as expected.

Do I misunderstand the command command? Or why doesn't it work as announced? How can I find out?

GNU bash, Version 5.1.16(1)-release (x86_64-pc-linux-gnu)
Copyright (C) 2020 Free Software Foundation, Inc.
command-line

1 Answers

  1. Best Answer

    It does work as announced: it suppresses shell function lookup, not built-in lookup. The effect with time is misleading: time is a reserved word, not a built-in, and only at the start of a pipeline. In command time, the pipeline is started by command, so the only time it could have executed was the external one.

    To see the difference, define a function:

    echo () { date; }

    And compare echo with command echo. (You can remove the function with unset -f echo.)

    To see what something is, use type:

    $ foo () { true; }
$ type if time [[ [ echo foo
if is a shell keyword
time is a shell keyword
[[ is a shell keyword
[ is a shell builtin
echo is a shell builtin
foo is a function
foo ()
{
    true
}
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