Question

I have some c code that I compile and run, in a directory that is accessible from many different unix computers (various linux and mac, occasionally others), with different OS’s obviously needing different executables.

I have a simple shell script that invokes the appropriate executable, prog.$OSTYPE.$MACHTYPE, compiling it first if necessary. This is very simple (although it requires using csh in order to have $OSTYPE and $MACHTYPE be reliably defined) and it almost works.

However, it turns out that even $OSTYPE and $MACHTYPE are not enough: for example, compiling on OSX 10.5 yields an executable prog.darwin.i386 which, when invoked on OSX 10.4, crashes instantly.

Yes, recompiling every time I want to run the program is one way to solve this, but it seems excessive. I know having a bin directory on every machine is a standard solution, but a non-root user may not have much write access outside their home directory (which is common to all the machines).

So my question is, is there a better approach? The compiler (often gcc) obviously knows what kind of system it is compiling for — is there a good portable way to find out what “kind of system” my script is running on, so it can invoke the correct executable, instead of one with undefined behavior?

Answer 1

You could use gcc -v to figure out what the installed/runnable gcc thinks is the target arch for hosted compiling (something like $(gcc -v 2>&1 | grep Target: | sed 's/.*: *//') in bash)

edit

If you really want to be able to do this without having anything in particular installed, you could extract the config.guess script from gcc (its in the top level directory of any gcc source package) and run that. Unfortunately this won’t work for all systems and might not exactly match what the system gcc package uses for some distributions, but this is the script used to configure gcc for building unless you explicitly override it…