I have a shared object ( libxyz.so ). Given LD_LIBRARY_PATH, how can find the exact location of this shared object? If i had a binary that depends on this lib, i would have used ldd on that.
Here is the reason why i ask:
I have a cgi script which works when using LD_LIBRARY_PATH set to say VALUE1. It does not work when the path is set to VALUE2. I would like to find the exact location of the library as specified by the path in VALUE1 ( Note that VALUE1 has almost 20+ different locations )
Platform: Linux
Put this in a file:
#!/bin/bash
IFS=:
for p in ${LD_LIBRARY_PATH}; do
if [ -e ${p}/libxyz.so ]; then
echo ${p}
fi
done
and run it.
You can also use ldd. To do this, you would:
Set LD_LIBRARY_PATH to the value when it works (i.e. export LD_LIBRARY_PATH=VALUE1)
Run ldd /path/to/prog | grep libxyz.so
put a sleep(30); in your cgi, launch it from a browser, then look into /proc/$(pidof mycgi)/maps for actual libs used by your program.
Related
I create a binary myBinary via cmake/CMakeLists.txt.
I would like to "include" default options on my binary.
In other words, I want my binary to be called with myBinary --option myopt even when I just run ./myBinary
How can I do that?
CMake does not have built-in support for you you want to do.
One solution is to do as #Youka said - change the source code of your program.
Another solution that I have used sometimes is to autogenerate a script that executes an executable:
# Create startup script
MACRO(GEN_START_SCRIPT binName)
# Generate content
SET(fileContent
"#!/bin/bash\n"
"\n"
"# This startup script is auto generated - do not modify!\n"
"\n"
"${binName} -a 23 -b 34 -c 976\n"
"\n"
)
# Write to file
SET(fileName ${CMAKE_CURRENT_BINARY_DIR}/${binName}.sh)
FILE(WRITE ${fileName} ${fileContent})
ENDMACRO()
Then call the macro after defining your executable:
ADD_EXECUTABLE(myBinary file1.c file.2)
GEN_START_SCRIPT(myBinary)
You can of course add other stuff to the script, like environment variables etc.
If you're in control of the sources and you want different default behavior... change the sources!
This is in no way a build system issue (CMake or otherwise).
I'm not sure why this isn't working, but perhaps I've oversimplified/overcomplicated things
I'm writing a Perl script that ultimately needs to call an external program. The catch is, this program needs a modified version of the LD_LIBRARY_PATH environment variable, in order to find a couple of libraries which the vendor does not install in standard places.
OK, the environment is in %ENV, which can be rewritten, yes?
I thought if I changed LD_LIBRARY_PATH in the parent, it would affect the dynamic linking of the child.
So I have:
use Env qw(#LD_LIBRARY_PATH);
use IPC::System::Simple qw(capturex $EXITVAL);
# We need these to establish the call to rsq later
my ($rsqexe, $rsqhome, $suffix) = fileparse($config->rsq());
push #LD_LIBRARY_PATH, $rsqhome;
eval {
$output = capturex(
$config->rsq(),
qq/"$source"/
);
};
But the child process dies with an error indicating the shared libraries can't be found.
How can I improve this?
I do need to examine the contents of $output after successful execution.
eval {
$output = capturex(
$config->rsq(),
qq/"$source"/
);
};
Here's the problem: I wasn't examining what was in $# (or $EVAL_ERR if you use ENGLISH;)
If I had, I would have seen that the problem was with the quoting qq/"$source"/ - because capturex() doesn't call the shell (which was the desired behaviour) the quotes break the file name (i.e. test.pdf exists, but ""test.pdf"" does not).
I have a environment variable set with name $MY_ENV_VARIABLE.
How do I use this variable inside my makefile to (for example) include some source files?
LOCAL_SRC_FILES = $(MY_ENV_VARIABLE)/libDEMO.so
Something like above doesn't seem to work.
Note: in my case this is needed for building with the Android NDK but I guess this applies to make in general.
Just to add some information...
The syntax to access the environment variable in make is like other variables in make...
#export the variable. e.g. in the terminal,
export MY_ENV_VARIABLE="hello world"
...
#in the makefile (replace before call)
echo $(MY_ENV_VARIABLE)
This performs the substitution before executing the commmand. If you instead, want the substitution to happen during the command execution, you need to escape the $ (For example, echo $MY_ENV_VARIABLE is incorrect and will attempt to substitute the variable M in make, and append it to Y_ENV_VARIABLE)...
#in the makefile (replace during call)
echo $$MY_ENV_VARIABLE
Make sure you exported the variable from your shell. Running:
echo $MY_ENV_VARIABLE
shows you whether it's set in your shell. But to know whether you've exported it so that subshells and other sub-commands (like make) can see it try running:
env | grep MY_ENV_VARIABLE
If it's not there, be sure to run export MY_ENV_VARIABLE before running make.
That's all you need to do: make automatically imports all environment variables as make variables when it starts up.
I just had a similar issue (under Cygwin):
Running echo $OSTYPE on the shell prints the value, but
running env | grep OSTYPE doesn't give any output.
As I can't guarantee that this variable is exported on all machines I want to run that makefile on, I used the following to get the variable from within the makefile:
OSTYPE = $(shell echo $$OSTYPE)
Which of course can also be used within a condition like the following:
ifeq ($(shell echo $$OSTYPE),cygwin)
# ...do something...
else
# ...do something else...
endif
EDIT:
Some things I found after experimenting with the info from jozxyqk's answer, all from within the makefile:
If I run #echo $$OSTYPE or #echo "$$OSTYPE" in a recipe, the variable is successfully expanded into cygwin.
However, using that in a condition like ifeq ($$OSTYPE,cygwin) or ifeq ("$$OSTYPE","cygwin") doesn't expand it.
Thus it is logical that first setting a variable like TEST = "$$OSTYPE" will lead to echo $(TEST) printing cygwin (the expansion is done by the echo call) but that doesn't work in a condition - ifeq ($(TEST),cygwin) is false.
I need to find out which library will be loaded given in the information returned from /sbin/ldconfig. I came up with the following:
#!/bin/bash
echo $(dirname $(/sbin/ldconfig -p | awk "/$1/ {print \$4}" | head -n 1))
Running this results with:
$ whichlib libGL.so
/usr/X11R6/lib
This a two part question:
Will this produce a reliable result across platform?
Is there a slicker way to parse the output of ldconfig?
Thanks,
Paul
There're several ways the library is loaded by executeable:
1.
Using $LD_LIBRARY_PATH
Using ld cache
Libary with full path compiled into binary (-rpath gcc flag)
You're using option 2, while option 1 and 3 are not considered.
Depending on what exactly you're doing you may want to run ldd directly on the executable you're planning to run rather than the general case ldconfig.
Since you asked, you could write your script like this:
dirname "$(/sbin/ldconfig -p | awk "\$1 == "$1" {print \$4; exit}")"
It's a little more precise and has one less pipe. Also echo $(cmd) is redundant; you can just write cmd.
According to g++ -print-search-dirs my C++ compiler is searching for libraries in many directories, including ...
/lib/../lib/:
/usr/lib/../lib/:
/lib/:
/usr/lib/
Naively, /lib/../lib/ would appear to be the same directory as /lib/ — lib's parent will have a child named lib, "that man's father's son is my father's son's son" and all that. The same holds for /usr/lib/../lib/ and /usr/lib/
Is there some reason, perhaps having to do with symbolic links, that g++ ought to be configured to search both /lib/../lib/ and /lib/?
If this is unnecessary redundancy, how would one go about fixing it?
If it matters, this was observed on an unmodified install of Ubuntu 9.04.
Edit: More information.
The results are from executing g++ -print-search-dirs with no other switches, from a bash shell.
Neither LIBRARY_PATH nor LPATH are output from printenv, and both echo $LPATH and echo LIBRARY_PATH return blank lines.
An attempt at an answer (which I gathered from a few minutes of looking at the gcc.c driver source and the Makefile environment).
These paths are constructed in runtime from:
GCC exec prefix (see GCC documentation on GCC_EXEC_PREFIX)
The $LIBRARY_PATH environment variable
The $LPATH environment variable (which is treated like $LIBRARY_PATH)
Any values passed to -B command-line switch
Standard executable prefixes (as specified during compilation time)
Tooldir prefix
The last one (tooldir prefix) is usually defined to be a relative path:
From gcc's Makefile.in
# Directory in which the compiler finds libraries etc.
libsubdir = $(libdir)/gcc/$(target_noncanonical)/$(version)
# Directory in which the compiler finds executables
libexecsubdir = $(libexecdir)/gcc/$(target_noncanonical)/$(version)
# Used to produce a relative $(gcc_tooldir) in gcc.o
unlibsubdir = ../../..
....
# These go as compilation flags, so they define the tooldir base prefix
# as ../../../../, and the one of the library search prefixes as ../../../
# These get PREFIX appended, and then machine for which gcc is built
# i.e i484-linux-gnu, to get something like:
# /usr/lib/gcc/i486-linux-gnu/4.2.3/../../../../i486-linux-gnu/lib/../lib/
DRIVER_DEFINES = \
-DSTANDARD_STARTFILE_PREFIX=\"$(unlibsubdir)/\" \
-DTOOLDIR_BASE_PREFIX=\"$(unlibsubdir)/../\" \
However, these are for compiler-version specific paths. Your examples are likely affected by the environment variables that I've listed above (LIBRARY_PATH, LPATH)
Well, theoretically, if /lib was a symlink to /drive2/foo, then /lib/../lib would point to /drive2/lib if I'm not mistaken. Theoretically...
Edit: I just tested and it's not the case - it comes back to /lib. Hrm :(