Say how to let a program A prints its binary location?
Thanks.
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Say how to let a program A prints its binary location?
Thanks.
There's no real standard way of doing that, but argv[0] contains the name of the application, and in most shell's the path is included - however, that's no guarantee, as it's the code calling exec() [or one of it's close relatives] that is responsible for filling in argv[0].
The only other option is to search in the path [which of course can also fail if the application was started with a path, but exec called without it, and the path does not contain the executable's directory].
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Mats
getcwd() on UNIX, GetCurrentDirectory() on Windows.Quote:
Originally Posted by meili100
On windows you can use GetModuleFileName().
That gives you the directory where the application is currently running, but not the location out of which the executable came.Quote:
Originally Posted by brewbuck
I agree that the question is not entirely clear. If I'm doing:
do you want to know the current directory (/home/me), or the /bin that is the executable /bin/ls's directory [and we're assuming that ls isn't built into the shell - which happens sometimes].Code:/home/me $ ls
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Mats
Hmm -- I just realized the question is unclear. The subject line says "working dir" but in the body of your post you ask for the binary's location. Which one of these are you actually interested in?
Does anybody know if there is a dirname() function?
Sorry for the misleading, I mean the binary dir.
In the "/home/me $ ls" case the output shall be /bin/ls
In that case, the solution is messy and somewhat platform dependent. The value of argv[0] is a relative or absolute path to the binary itself. If it is absolute, then you already have your answer. If it is relative, then it is relative to the current directory, so you must get the curr dir and then walk upward until you reach the root of the filesystem to construct the full path.Quote:
Originally Posted by meili100
And since argv[0] is provided by the caller to exec, not by the OS when it found the executable file, it's by no means guaranteed that the argv[0] is actually containing the path to the executable.Quote:
Originally Posted by brewbuck
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Mats
I found /proc/self/exe is linked to the binary location
for example :
/proc/self/exe -> /bin/ls
But the question how to get the "/bin/ls" string?
A read from /proc/self/exe perhaps?
Of coruse, if you want to do that with /bin/ls, then you need to recompile the source of /bin/ls - if you want to find '/home/me/myproject/bin/myprog', then you just need to open and read /proc/self/exe in your "myprog.c" or whatever your source file is.
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Mats
Thanks, but
is a link.Code:/proc/self/exe -> /home/me/myproject/bin/myprog
How to get the
Code:/home/me/myproject/bin/myprog
Very true, although in most cases it will be correct.Quote:
Originally Posted by matsp
You could always parse ls -l or something . . . .Quote:
Originally Posted by meili100
. . . or, if you can convince "open" or whatever not to follow symbolic links, you could just read the file directly.
Ok, so you probably need to use opendir()/readdir()/closedir() to open/read/close the directory entry then. Unfortunately, you won't be able to spawn a process and use something like a "ls | awk ..." or "ls | sed ...", since you would then get the path to "ls", rather than your application, which I assume is what you are actually after [although I suppose the stuff is also in a numbered process /proc/<pid>/exe (or similar) directory, so I suppose that would be possible to use for such scripting tricks].Quote:
Originally Posted by meili100
Edit: Note that you would have to do (pseudo-code)Code:x = opendir("\proc\self");
while(readdir(dirinfo))
{
if (dirinfo.fname == "exe")
{
getlinkcontent();
break;
}
}
closedir(x);
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Mats