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makefileopensslrpath

How to compile OpenSSL with relative rpath

I have been trying to compile openssl 1.0.0g with the following rpath:

$ORIGIN/../lib64

Everytime I readelf -d apps/openssl, I am getting results like the following depending on what escaping variation I tried:

\RIGIN/../lib64
RIGIN/../lib64
ORIGIN/../lib64

I want to setup my rpath without using external tools like chrpath. Is it at all possible? I will basically accept anything that does not involve using external tools like chrpath (though I would already be done with that).

Ideally, I would like to do it by passing options on the command line (any form of -Wl,-rpath,$ORIGIN/../lib64).

I don't mind editing the generated Makefile, which is what I have been trying last. If only I could get it to print a stupid dollar sign!!! I tried modifying LIBRPATH under the BUILDENV= block with no luck. My best results so far:

LIBRPATH=$$'ORIGIN/../lib64 # result: /../lib64
LIBRPATH=$$$$'ORIGIN/../lib64 # result: 12345<pid>/../lib64

I have read various rpath related questions and tried various escaping and quoting tricks but nothing worked so far!

Solution

  • I went the chrpath way. http://enchildfone.wordpress.com/2010/03/23/a-description-of-rpath-origin-ld_library_path-and-portable-linux-binaries/

    It is quite complicated to counter shell expansion of `$$ORIGIN`` in openssl. Sooner or later, it gets expanded because of the dollar sign. If you really want to go this way, you can do it. I have found the following to work with openssl 1.0.1g on Linux. In Makefile.shared, look for this line:

    DO_GNU_APP=LDFLAGS="$(CFLAGS) -Wl,-rpath,$(LIBRPATH)"

    Replace it with the following. This quoting-fu neutralize the expansion of $. The double $$ is the way to get a single dollar sign in makefiles.

    DO_GNU_APP=LDFLAGS="$(CFLAGS) -Wl,-rpath,'"'$$'"ORIGIN/../lib64'"

    After compiling:

    readelf -d apps/openssl | grep RPATH
 0x000000000000000f (RPATH)              Library rpath: ['$ORIGIN/../lib64']