How to capture "*" as argument to bash function and use in a comparison

Question

I want to have bash run rm -i * if I type rm * but otherwise run regular rm. Importantly, I don't want to run rm -i every time I use a wildcard such as rm part*. Here is as far as I could come up with:

rm ()
{
    if [ "$1" == "*" ]; then
        rm -i *
    else
        rm $1
    fi
}

But I know this will fail. I know that the comparison I want is to ^*$, but I don't know how to implement it.

Answer 1

It's literally impossible to know if your command was called with a wildcard without the cooperation of your shell.

When you invoke rm * (like any other command), the * is replaced with a list of filenames before invocation. Thus, when inside the command, the information that it was given a wildcard no longer exists: $1, $2, etc. have been replaced with a list of names that the wildcard expanded to.

---

That said, since we're a shell function, the cooperation of our shell is actually available:

rm() {
  local cmd
  read -r _ cmd < <(HISTTIMEFORMAT=''; history 1)
  if [[ $cmd = "rm *" ]]; then
    command rm -i "$@"
  else
    command rm "$@"
  fi
}

---

How does this work?

• history 1 returns the most recent command in the shell's history (preceded by a number).
• read -r _ cmd reads that number into the variable _, and the rest of the command line into the variable cmd
• [[ $cmd = "rm *" ]] compares the command against that precise string
• command rm ... runs the external rm command, avoiding recursion back to our function again.