Run a command that is shadowed by an alias
Let’s say I have the following alias in bash – alias ls='ls --color=auto'
– and I want to call ordinary ls
without options. Is the only way to do that is to unalias, do the command and then alias again? Or there is some nifty trick or workaround?
The alias is just a defined shortcut. In this example, the alias defined is the string ls
, which executes ls --color=auto
. If you don’t want that behavior, you can call the binary ls
using the absolute path.
So executing just /bin/ls
will produce output without color because it is not the alias you defined.
You could also change the alias to something different, or add a new alias that executes just ls
without the color parameters.
You can also prefix a back slash to disable the alias: ls
Edit: Other ways of doing the same include:
Use “command”: command ls
as per Mikel.
Use the full path: /bin/ls
as per uther.
Quote the command: "ls"
or 'ls'
as per Mikel comment.
You can remove the alias temporarily for that terminal session with unalias command_name
.
Typing the command in uppercase seems to work, like LS
, though I’m not really sure why.
Another way, building upon @uther’s answer, is to use $(which ...)
or `which ...`
(using backticks ``
) to automatically find the executable’s path. This will work in bash
:
$(which ls)
`which ls`
Undoubtedly, simply prefixing the command with a is much shorter.
UPDATE: According to How to use `which` on an aliased command?, this doesn’t seem to be reliable at all.
Personally, I just avoid defining aliases with the same names as commands, so ls
always invokes the actual command. I have aliases or functions for various combinations of options like l
, ll
, and lg
. (This means I need unalias ls
in my .bashrc
or .cshrc
.)
I find that the OS’s assumptions about which options I might prefer (overriding the assumptions of the designers of the ls
command itself) rarely match my own personal tastes. I happen to dislike the look of ls --color=auto
, and its legibility can vary greatly between black-on-white and white-on-black.
YMMV, and the other solutions are of course still good to know.
If ls
and command ls
are "not working", use type -a ls
to troubleshoot what is happening.
In my case, ls
and command ls
were returning nothing, and I was going crazy trying to figure out why. Turns out I somehow had a zero-byte file named ls
in one of my paths, ~/bin
.
I was able to fix the situation by deleting ~/bin/ls
. Then hash -d ls
to remove it from the hash table.
Side note:
which -a ls
was not helpful because it only returned /bin/ls
. Apparently, which
cannot interpret the tilde in my path: ~/bin
. However type -a ls
can.
For zsh
Order:
- you type in the shell (stdin)
- history expansion, such as
!!
,!-3
- alias expansion (try ctrl–x then a, to see the expanded result)
- other expansions (order: Process Substitution > Parameter Expansion > Command Substitution > Arithmetic Expansion > Brace Expansion)
suppose we have:
alias pwd='echo "An Alias"'
then this function definition will get the error zsh: defining function based on alias 'pwd'
:
pwd(){
echo "A function"
}
- alias expansion may be avoided by adding any kind of quoting to the word. If you want to use unaliased
pwd
, type'pwd'
,pwd
or"pwd"
(`pwd` is not regarded as quoting) builtin pwd
use zsh’ s built-in function(aka command?)pwd
command pwd
use /bin/pwd, (from GNU)
If there is no alias pwd="echo WhatEver_blabla"
, and this is in zshrc:
pwd(){
echo "A function"
}
then quoting like 'pwd'
still calls this function.
But command pwd
and built pwd
work as expected
So we can have alias rg="rg --hiddeng"
, but not rg(){ rg --hidden}
(Otherwise we will meet such error : maximum nested function level reached; increase FUNCNEST?)
ref:
man zshall