Bash function calling command given as argument

Question

How do you write a function in bash that executes the command that it is given as an argument, where

• The given command may be an alias
• Arguments must be passed on exactly as given; no evaluating may be done

In other words, how to write an as-transparent-as-possible wrapper function.

The goal of the wrapper function could for example be to set the current directory before and after the given command, and/or set environment variables, or time how long the given command takes,... As a simple example here I take a function that just prints a line and then executes the given command.

A first attempt:

function wrap1 {
   echo Starting: "$@"
   "$@"
}

You could use it like wrap1 echo hello. But the problem is you cannot do alias myalias echo and then call wrap1 myalias hello: it wouldn't resolve the alias.

Another attempt using eval:

function wrap2 {
   echo Starting: "$@"
   eval "$@"
}

Now calling an alias works. But the problem is it evaluates the arguments too. For example wrap2 echo "\\a" prints just a instead of \a because the arguments are evaluated twice.

shopt -s expand_aliases doesn't seem to help here either.

Is there a way to both evaluate aliases like wrap2, but still pass on the arguments directly like wrap1?

Answer 1

It seems to be possible with a double eval:

eval "eval x=($(alias y | cut -s -d '=' -f 2))"
# now the array x contains the split expansion of alias y
"${x[@]}" "${other_args[@]}"

So maybe your function could be written as follows:

wrap() {
    eval "eval prefix=($(alias $1 | cut -s -d '=' -f 2))"
    shift
    "${prefix[@]}" "$@"
}

However, eval is evil, and double eval is double evil, and aliases are not expanded in scripts for a reason.

Answer 2

You (uh, I) can use printf %q to escape the arguments. At first sight, escaping with printf and then doing eval always gives the same result as passing the arguments directly.

wrap() {
    echo Starting: "$@"
    eval $(printf "%q " "$@")
}