$@ in alias inside script: is there a "local" $@?

Question

I have aliased pushd in my bash shell as follows so that it suppresses output:

alias pushd='pushd "$@" > /dev/null'

This works fine most of the time, but I'm running into trouble now using it inside functions that take arguments. For example,

test() { pushd . ... }

Running test without arguments is fine. But with arguments:

> test x y z bash: pushd: too many arguments

I take it that pushd is trying to take . x y z as arguments instead of just .. How can I prevent this? Is there a "local" equivalent of $@ that would only see . and not x y z?

@Wildcard Because I had copy/pasted that line from someone else who was apparently also a beginner like me. :S — Théophile
– Théophile, Commented Aug 16, 2019 at 15:00

Stéphane Chazelas · Accepted Answer · 2019-08-15 18:57:29Z

Aliases define a way to replace a shell token with some string before the shell event tries to parse code. It's not a programming structure like a function.

In

alias pushd='pushd "$@" > /dev/null'

and then:

pushd .

What's going on is that the pushd is replaced with pushd "$@" > /dev/null and then the result parsed. So the shell ends up parsing:

pushd "$@" > /dev/null .

Redirections can appear anywhere on the command line, so it's exactly the same as:

pushd "$@" . > /dev/null

or

> /dev/null pushd "$@" .

When you're running that from the prompt, "$@" is the list of arguments your shell received so unless you ran set arg1 arg2, that will likely be empty, so it will be the same as

pushd . > /dev/null

But within a function, that "$@" will be the arguments of the function.

Here, you either want to define pushd as a function like:

pushd() { command pushd "$@" > /dev/null; }

Or an alias like:

alias pushd='> /dev/null pushd'

or

alias pushd='pushd > /dev/null

I had actually thought of writing it as a function, but I didn't know about command yet, so of course it created an infinite loop. — Théophile
– Théophile, Commented Aug 15, 2019 at 21:07

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