How to intercept and remove a command line argument in bash

arguments, bash, command-line-arguments, shell

Solution

This should work:

# Run older ld (pseudo condition)
if [[ <old_ld_condition> ]]; then
    args=()
    for var; do
        # Ignore known bad arguments
        [[ $var != '-dependency_info' ]] && args+=("$var")
    done

    /path/to/old/ld "${args[@]}"
else
    /path/to/new/ld "$@"
fi

Problem

After "upgrading" to Mavericks and Xcode 5, I have a variety of minor problems to deal with to make Xcode compile some of my older projects. It appears that Xcode is passing a new argument to the `ld` linker, and there's really no stopping Xcode from doing so. An older version of `ld`, which I need for a variety of reasons, gives an error when seeing an argument it doesn't know (so my projects cannot compile). What I need is a thin wrapper over my older version of `ld` to remove the "bad" arguments under certain circumstances. I thought that a bash shell script would be perfect, but bash is not my forte. Here's what I've got: ``` # Look for conditions necessary to use older ld ... # (placeholder, obviously) # Run older ld (pseudo condition) if [ <old_ld_condition> ]; then ARGS='' for var in "$@"; do # Ignore known bad arguments if [ "$var" = '-dependency_info' ]; then continue fi ARGS="$ARGS $var" done /path/to/old/ld "$ARGS" else /path/to/new/ld "$@" fi ``` However, running `/path/to/old/ld "$ARGS"` results in `ld` interpreting the entire `$ARGS` string as one argument. Running `/path/to/old/ld $ARGS` results in `ld` receiving unescaped versions of previously escaped strings. Clearly, I'm misunderstanding something about the nature of `$@`, how to manipulate it, and how to pass that manipulation to the older `ld`. Thanks everyone.

Original source