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

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

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Wouldn't the first line be better written as "Why do we use ./command_name to execute a command in linux?" –  user15760 Jul 26 '13 at 6:18

4 Answers 4

up vote 20 down vote accepted

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."

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It is very common to execute programs in the current directory. Why doesn't the shell search in there as well? It first searches in ., then in $PATH. –  Michael Jul 9 '13 at 22:14

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
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Thanks.My doubt is why do you need ./ at the start.... I got the use of "." (to poit the current directory) but why "/" after that? –  Renjith G Nov 30 '10 at 8:35
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/ is the path separator in Linux, so you use it to separate the directory (.) from the filename (a.out). Without it you have .a.out which is a valid filename in its own right. (Try touch .a.out; ls -lA to see this.) –  Simon Whitaker Nov 30 '10 at 8:37
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that is how you specify the path in Unix, <dir>/<file> so you are basically saying execute a file in the current directory, which is indicated by ./test –  bronzebeard Nov 30 '10 at 8:37
    
Red Hat Linux release 9 (Shrike) Kernel 2.4.20-8 on an i686 [renjithg@cvsserver renjithg]$ touch .a.out;ls -lA total 3 -rw-rw-r-- 1 renjithg renjithg 0 Nov 30 13:46 .a.out -rwxrwxr-x 1 renjithg renjithg 11669 Nov 30 13:46 a.out -rw-rw-r-- 1 renjithg renjithg 218 Aug 24 2009 aray.c [renjithg@cvsserver renjithg]$ –  Renjith G Nov 30 '10 at 8:48
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Red Hat Linux 9? Time to upgrade! –  mattdm Feb 17 '11 at 1:53

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.

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+1 for why! Good explanation there. –  Josh Dec 1 '10 at 20:24
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A practical example of someone getting bit by this:unix.stackexchange.com/questions/7767 –  mattdm Feb 23 '11 at 20:13
    
what could go wrong with having . at the end? –  balki May 3 '11 at 12:57
    
Yeah, what "other attacks" are there? Apart from typos like "ls-l". –  aib Jul 4 '11 at 3:03
    
Typos, or else using a common command which sysadmins type often but which may not be installed — vim, for example. –  mattdm Oct 8 '12 at 13:11

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.

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