$@ 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?

Answer 1

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