Why do we use “./” to execute a file?

Question

Why do we use ./filename to execute a file in linux?

Why not just enter it like other commands gcc, ls etc...

Answer 1

You need the ./ bit to tell the shell where the executable is, since the current directory is unlikely to be in $PATH.

You can use which to get the full path to "other commands."

Answer 2

You can try to add :. to your $PATH variable.

Try ALT+F2 and type: gksudo gedit /etc/environment if running Linux/GTK (this is what you have if using Ubuntu).

HOWEVER, I strongly advise you NOT to do that. It's bad bad bad and bad.

You know, that kind of things work like this since 1970. There is a reason why the current directory isn't included in the $PATH.

. is the current directory

.something would be a hidden file (Type "ALT+" to make them appear in Nautilus, or try "ls -la".

./someProgram.sh is what you type to RUN an executable someProgram.sh in the current directory.

.somethingElse would mean that you have a hidden executable in the current directory, which is a bad idea.

Answer 3

The literal answer is as others have given: because the current directory isn't in your $PATH.

But why? In short, it's for security. If you're looking in someone else's home directory (or /tmp), and type just gcc or ls, you want to know you're running the real one, not a malicious version your prankster friend has written which erases all your files. Having . as the last entry in your path is a bit safer, but there are other attacks which make use of that.

Answer 4

If you mean, why do you need ./ at the start - that's because (unlike in Windows), the current directory isn't part of your path by default. If you run:

$ ls

your shell looks for ls in the directories in your PATH environment variable (echo $PATH to see it), and runs the first executable called ls that it finds. If you type:

$ a.out

the shell will do likewise - but it probably won't find an executable called a.out. You need to tell the shell where a.out is - it it's in the current directory (.) then the path is ./a.out.

If you're asking why it's called "a.out", that's just the default output file name for gcc. You can change it with the -o command line arg. For example:

$ gcc test.c -o test
$ ./test