How to intercept and remove a command line argument in bash

Question

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.

You can select the command-line toolset version from within Xcode, I do not know if that will help you or not? Generally unless you did a complete fresh install, the old SDKs and command-line toolset versions will coexist. — Andon M. Coleman
– Andon M. Coleman, Commented Feb 4, 2014 at 2:52
These projects are actual Xcode .xcodeproj files, and unfortunately I'm not in the position to bring them over to a command-line build process (e.g. make). Thanks for the suggestion though. — inspector-g
– inspector-g, Commented Feb 4, 2014 at 5:29

anubhava · Accepted Answer · 2022-07-27 09:24:25Z

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

user3159253 · Accepted Answer · 2014-02-04 03:27:34Z

You should use Bash Arrays if you really want to stay with bash:

declare -a ARGS for var in "$@"; do # Ignore known bad arguments if [ "$var" = '-dependency_info' ]; then continue fi ARGS[${#ARGS[@]}]="$var" done

now "${ARGS[@]}" can be used just as "$@". man bash for more information.

Simon Schröter · Accepted Answer · 2024-09-03 13:17:10Z

So what about having this?

#!/bin/bash if [ <old_ld_condition> ]; then /path/to/old/ld "${@//-dependency_info/}" else /path/to/new/ld "$@" fi

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