exec() error when using find -exec

Question

I've to find with a script an executable named Test that take as argument a path, and execute it. I'm doing this with this line:

find -name Test -exec {} path \;

In Test I got an execl:

    execl("./Test1","Test1",(char*)0);
    perror("Exec failed");
    exit(EXIT_FAILURE);

where Test1 is in the same directory of Test . Executing Test "manually" everything goes fine, but using the line written above I have a Exec failed: No such file or directory error.

What's wrong ?

Answer 1

find executes Test from the directory you are executing find. If you can change the code for Test, then put the absolute path of Test1:

execl("/home/myuser/some/path/Test1","Test1",(char*)0);
perror("Exec failed");
exit(EXIT_FAILURE);

Or you can use -execdir instead of -exec:

find -name Test -execdir {} path \;

From find manpage:

-execdir: Like -exec, but the specified command is run from the subdirectory containing the matched file, which is not normally the directory in which you started find