CODEWITHSUNDEEP
c++linuxeclipse
vico
vico

Reputation: 18171

Show environment variable LD_LIBRARY_PATH

I have simple console application that prints environment variables:

int main(int argc, char **argv, char **envp)
{

  printf("Scanning for LD_LIB_PATH\n");
  for (char **env = envp; *env != 0; env++)
  {
    char *thisEnv = *env;
    std::string s= *env;
    if (s.find("LD_LIBRARY_PATH")!=std::string::npos)
        {
        printf("!!!!!!! %s\n", thisEnv);
        }

    printf("%s\n", thisEnv);
  }

}

Before run executable I run script that sets LD_LIBRARY_PATH

export LD_LIBRARY_PATH=~/aaa/bbb/Debug:~/ccc/ddd/Debug
echo "searching:"
export | grep LD_LIBRARY
echo "done"

Script run fine with output:

exporting
searching:
declare -x LD_LIBRARY_PATH="/home/vicp/aaa/bbb/Debug:/home/vico/ccc/ddd/Debug"
done

I run executable and it finds many variables but no environment variable LD_LIB_PATH. Why?

UPD

As recommended I script . ./script.sh Then double check with command:

export |grep LD_LIBRARY_PATH

Got output:

declare -x LD_LIBRARY_PATH="/home/vicp/aaa/bbb/Debug:/home/vico/ccc/ddd/Debug"

But still don't see LD_LIBRARY_PATH in my programm.

Upvotes: 0

Views: 4310

Answers (3)

awahl
awahl

Reputation: 24

It has nothing to do with your C/C++ application. try the following: $ ./script.sh $ echo $LD_LIBRARY_PATH And you'll see that LD_LIBRARY_PATH is not set

When you launch your script, bash creates a new process with its environment inherited from the original bash process. In that newly created process, you set the process environment variable LD_LIBRARY_PATH=xxxx When finalized your script.sh exits, and its environment dies with it.

Meaning the LD_LIBRARY_PATH is not set in your original shell.

As mentioned here above you need to run your script in the current shell Either with . or with source.

I tested with your C/C++ and it works

Upvotes: 0

Gunnar Sigfusson
Gunnar Sigfusson

Reputation: 81

Look at the line printf(!!!!!!!"%s\n", thisEnv); Change to printf("!!!! %s\n", thisEnv);

Upvotes: 0

mmlr
mmlr

Reputation: 2155

Depending on how you run the script, the env variable will only be added to the environment of the subshell running the script.

If you run the script like this:

$ ./script.sh

This will spawn a new shell wherein the script is run. The parent shell, i.e. the one you started the script from, will not be affected by what the script does. Changes to the environment will therefore not be visible (changing the working directory or similar will also not work).

If the script is intended to modify the environment of the current shell, the script needs to be sourced using the . or source commands:

$ . ./script.sh
$ source ./script.sh

From help source in bash:

Execute commands from a file in the current shell.

Then there seems to be a problem with the code:

As a commenter stated previousy, there are a lot of exclamation points in the success case printf. I presume these were meant to highlight the found variable name. Since they are outside of the opening quote of the format string, they are interpreted as logical not operators.

The result is that the format string literal (a pointer) is negated, resulting in false due to the number of operators. This usually maps to 0, which would mean the format string becomes a null pointer. What happens when handing a null pointer to printf as the format string depends, as far as I can tell, on the implementation. If it doesn't crash, it most certainly will not print anything however.

So the code may actually work, but there is no visible output due to that bug. To fix it, move the exclamation points into the format string, i.e. after the opening quote.

Upvotes: 3

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