When your executable behaves differently when invoked with ./ [duplicate] - c

When running scripts in bash, I have to write ./ in the beginning:
$ ./manage.py syncdb
If I don't, I get an error message:
$ manage.py syncdb
-bash: manage.py: command not found
What is the reason for this? I thought . is an alias for current folder, and therefore these two calls should be equivalent.
I also don't understand why I don't need ./ when running applications, such as:
user:/home/user$ cd /usr/bin
user:/usr/bin$ git
(which runs without ./)

Because on Unix, usually, the current directory is not in $PATH.
When you type a command the shell looks up a list of directories, as specified by the PATH variable. The current directory is not in that list.
The reason for not having the current directory on that list is security.
Let's say you're root and go into another user's directory and type sl instead of ls. If the current directory is in PATH, the shell will try to execute the sl program in that directory (since there is no other sl program). That sl program might be malicious.
It works with ./ because POSIX specifies that a command name that contain a / will be used as a filename directly, suppressing a search in $PATH. You could have used full path for the exact same effect, but ./ is shorter and easier to write.
EDIT
That sl part was just an example. The directories in PATH are searched sequentially and when a match is made that program is executed. So, depending on how PATH looks, typing a normal command may or may not be enough to run the program in the current directory.

When bash interprets the command line, it looks for commands in locations described in the environment variable $PATH. To see it type:
echo $PATH
You will have some paths separated by colons. As you will see the current path . is usually not in $PATH. So Bash cannot find your command if it is in the current directory. You can change it by having:
PATH=$PATH:.
This line adds the current directory in $PATH so you can do:
manage.py syncdb
It is not recommended as it has security issue, plus you can have weird behaviours, as . varies upon the directory you are in :)
Avoid:
PATH=.:$PATH
As you can “mask” some standard command and open the door to security breach :)
Just my two cents.

Your script, when in your home directory will not be found when the shell looks at the $PATH environment variable to find your script.
The ./ says 'look in the current directory for my script rather than looking at all the directories specified in $PATH'.

When you include the '.' you are essentially giving the "full path" to the executable bash script, so your shell does not need to check your PATH variable. Without the '.' your shell will look in your PATH variable (which you can see by running echo $PATH to see if the command you typed lives in any of the folders on your PATH. If it doesn't (as is the case with manage.py) it says it can't find the file. It is considered bad practice to include the current directory on your PATH, which is explained reasonably well here: http://www.faqs.org/faqs/unix-faq/faq/part2/section-13.html

On *nix, unlike Windows, the current directory is usually not in your $PATH variable. So the current directory is not searched when executing commands. You don't need ./ for running applications because these applications are in your $PATH; most likely they are in /bin or /usr/bin.

This question already has some awesome answers, but I wanted to add that, if your executable is on the PATH, and you get very different outputs when you run
./executable
to the ones you get if you run
executable
(let's say you run into error messages with the one and not the other), then the problem could be that you have two different versions of the executable on your machine: one on the path, and the other not.
Check this by running
which executable
and
whereis executable
It fixed my issues...I had three versions of the executable, only one of which was compiled correctly for the environment.

Rationale for the / POSIX PATH rule
The rule was mentioned at: Why do you need ./ (dot-slash) before executable or script name to run it in bash? but I would like to explain why I think that is a good design in more detail.
First, an explicit full version of the rule is:
if the path contains / (e.g. ./someprog, /bin/someprog, ./bin/someprog): CWD is used and PATH isn't
if the path does not contain / (e.g. someprog): PATH is used and CWD isn't
Now, suppose that running:
someprog
would search:
relative to CWD first
relative to PATH after
Then, if you wanted to run /bin/someprog from your distro, and you did:
someprog
it would sometimes work, but others it would fail, because you might be in a directory that contains another unrelated someprog program.
Therefore, you would soon learn that this is not reliable, and you would end up always using absolute paths when you want to use PATH, therefore defeating the purpose of PATH.
This is also why having relative paths in your PATH is a really bad idea. I'm looking at you, node_modules/bin.
Conversely, suppose that running:
./someprog
Would search:
relative to PATH first
relative to CWD after
Then, if you just downloaded a script someprog from a git repository and wanted to run it from CWD, you would never be sure that this is the actual program that would run, because maybe your distro has a:
/bin/someprog
which is in you PATH from some package you installed after drinking too much after Christmas last year.
Therefore, once again, you would be forced to always run local scripts relative to CWD with full paths to know what you are running:
"$(pwd)/someprog"
which would be extremely annoying as well.
Another rule that you might be tempted to come up with would be:
relative paths use only PATH, absolute paths only CWD
but once again this forces users to always use absolute paths for non-PATH scripts with "$(pwd)/someprog".
The / path search rule offers a simple to remember solution to the about problem:
slash: don't use PATH
no slash: only use PATH
which makes it super easy to always know what you are running, by relying on the fact that files in the current directory can be expressed either as ./somefile or somefile, and so it gives special meaning to one of them.
Sometimes, is slightly annoying that you cannot search for some/prog relative to PATH, but I don't see a saner solution to this.

When the script is not in the Path its required to do so. For more info read http://www.tldp.org/LDP/Bash-Beginners-Guide/html/sect_02_01.html

All has great answer on the question, and yes this is only applicable when running it on the current directory not unless you include the absolute path. See my samples below.
Also, the (dot-slash) made sense to me when I've the command on the child folder tmp2 (/tmp/tmp2) and it uses (double dot-slash).
SAMPLE:
[fifiip-172-31-17-12 tmp]$ ./StackO.sh
Hello Stack Overflow
[fifi#ip-172-31-17-12 tmp]$ /tmp/StackO.sh
Hello Stack Overflow
[fifi#ip-172-31-17-12 tmp]$ mkdir tmp2
[fifi#ip-172-31-17-12 tmp]$ cd tmp2/
[fifi#ip-172-31-17-12 tmp2]$ ../StackO.sh
Hello Stack Overflow

Related

Execute compiled executable C file without ./ [duplicate]

When running scripts in bash, I have to write ./ in the beginning:
$ ./manage.py syncdb
If I don't, I get an error message:
$ manage.py syncdb
-bash: manage.py: command not found
What is the reason for this? I thought . is an alias for current folder, and therefore these two calls should be equivalent.
I also don't understand why I don't need ./ when running applications, such as:
user:/home/user$ cd /usr/bin
user:/usr/bin$ git
(which runs without ./)
Because on Unix, usually, the current directory is not in $PATH.
When you type a command the shell looks up a list of directories, as specified by the PATH variable. The current directory is not in that list.
The reason for not having the current directory on that list is security.
Let's say you're root and go into another user's directory and type sl instead of ls. If the current directory is in PATH, the shell will try to execute the sl program in that directory (since there is no other sl program). That sl program might be malicious.
It works with ./ because POSIX specifies that a command name that contain a / will be used as a filename directly, suppressing a search in $PATH. You could have used full path for the exact same effect, but ./ is shorter and easier to write.
EDIT
That sl part was just an example. The directories in PATH are searched sequentially and when a match is made that program is executed. So, depending on how PATH looks, typing a normal command may or may not be enough to run the program in the current directory.
When bash interprets the command line, it looks for commands in locations described in the environment variable $PATH. To see it type:
echo $PATH
You will have some paths separated by colons. As you will see the current path . is usually not in $PATH. So Bash cannot find your command if it is in the current directory. You can change it by having:
PATH=$PATH:.
This line adds the current directory in $PATH so you can do:
manage.py syncdb
It is not recommended as it has security issue, plus you can have weird behaviours, as . varies upon the directory you are in :)
Avoid:
PATH=.:$PATH
As you can “mask” some standard command and open the door to security breach :)
Just my two cents.
Your script, when in your home directory will not be found when the shell looks at the $PATH environment variable to find your script.
The ./ says 'look in the current directory for my script rather than looking at all the directories specified in $PATH'.
When you include the '.' you are essentially giving the "full path" to the executable bash script, so your shell does not need to check your PATH variable. Without the '.' your shell will look in your PATH variable (which you can see by running echo $PATH to see if the command you typed lives in any of the folders on your PATH. If it doesn't (as is the case with manage.py) it says it can't find the file. It is considered bad practice to include the current directory on your PATH, which is explained reasonably well here: http://www.faqs.org/faqs/unix-faq/faq/part2/section-13.html
On *nix, unlike Windows, the current directory is usually not in your $PATH variable. So the current directory is not searched when executing commands. You don't need ./ for running applications because these applications are in your $PATH; most likely they are in /bin or /usr/bin.
This question already has some awesome answers, but I wanted to add that, if your executable is on the PATH, and you get very different outputs when you run
./executable
to the ones you get if you run
executable
(let's say you run into error messages with the one and not the other), then the problem could be that you have two different versions of the executable on your machine: one on the path, and the other not.
Check this by running
which executable
and
whereis executable
It fixed my issues...I had three versions of the executable, only one of which was compiled correctly for the environment.
Rationale for the / POSIX PATH rule
The rule was mentioned at: Why do you need ./ (dot-slash) before executable or script name to run it in bash? but I would like to explain why I think that is a good design in more detail.
First, an explicit full version of the rule is:
if the path contains / (e.g. ./someprog, /bin/someprog, ./bin/someprog): CWD is used and PATH isn't
if the path does not contain / (e.g. someprog): PATH is used and CWD isn't
Now, suppose that running:
someprog
would search:
relative to CWD first
relative to PATH after
Then, if you wanted to run /bin/someprog from your distro, and you did:
someprog
it would sometimes work, but others it would fail, because you might be in a directory that contains another unrelated someprog program.
Therefore, you would soon learn that this is not reliable, and you would end up always using absolute paths when you want to use PATH, therefore defeating the purpose of PATH.
This is also why having relative paths in your PATH is a really bad idea. I'm looking at you, node_modules/bin.
Conversely, suppose that running:
./someprog
Would search:
relative to PATH first
relative to CWD after
Then, if you just downloaded a script someprog from a git repository and wanted to run it from CWD, you would never be sure that this is the actual program that would run, because maybe your distro has a:
/bin/someprog
which is in you PATH from some package you installed after drinking too much after Christmas last year.
Therefore, once again, you would be forced to always run local scripts relative to CWD with full paths to know what you are running:
"$(pwd)/someprog"
which would be extremely annoying as well.
Another rule that you might be tempted to come up with would be:
relative paths use only PATH, absolute paths only CWD
but once again this forces users to always use absolute paths for non-PATH scripts with "$(pwd)/someprog".
The / path search rule offers a simple to remember solution to the about problem:
slash: don't use PATH
no slash: only use PATH
which makes it super easy to always know what you are running, by relying on the fact that files in the current directory can be expressed either as ./somefile or somefile, and so it gives special meaning to one of them.
Sometimes, is slightly annoying that you cannot search for some/prog relative to PATH, but I don't see a saner solution to this.
When the script is not in the Path its required to do so. For more info read http://www.tldp.org/LDP/Bash-Beginners-Guide/html/sect_02_01.html
All has great answer on the question, and yes this is only applicable when running it on the current directory not unless you include the absolute path. See my samples below.
Also, the (dot-slash) made sense to me when I've the command on the child folder tmp2 (/tmp/tmp2) and it uses (double dot-slash).
SAMPLE:
[fifiip-172-31-17-12 tmp]$ ./StackO.sh
Hello Stack Overflow
[fifi#ip-172-31-17-12 tmp]$ /tmp/StackO.sh
Hello Stack Overflow
[fifi#ip-172-31-17-12 tmp]$ mkdir tmp2
[fifi#ip-172-31-17-12 tmp]$ cd tmp2/
[fifi#ip-172-31-17-12 tmp2]$ ../StackO.sh
Hello Stack Overflow

"Command not found" trying to run my C program by typing its name at the terminal [duplicate]

When running scripts in bash, I have to write ./ in the beginning:
$ ./manage.py syncdb
If I don't, I get an error message:
$ manage.py syncdb
-bash: manage.py: command not found
What is the reason for this? I thought . is an alias for current folder, and therefore these two calls should be equivalent.
I also don't understand why I don't need ./ when running applications, such as:
user:/home/user$ cd /usr/bin
user:/usr/bin$ git
(which runs without ./)
Because on Unix, usually, the current directory is not in $PATH.
When you type a command the shell looks up a list of directories, as specified by the PATH variable. The current directory is not in that list.
The reason for not having the current directory on that list is security.
Let's say you're root and go into another user's directory and type sl instead of ls. If the current directory is in PATH, the shell will try to execute the sl program in that directory (since there is no other sl program). That sl program might be malicious.
It works with ./ because POSIX specifies that a command name that contain a / will be used as a filename directly, suppressing a search in $PATH. You could have used full path for the exact same effect, but ./ is shorter and easier to write.
EDIT
That sl part was just an example. The directories in PATH are searched sequentially and when a match is made that program is executed. So, depending on how PATH looks, typing a normal command may or may not be enough to run the program in the current directory.
When bash interprets the command line, it looks for commands in locations described in the environment variable $PATH. To see it type:
echo $PATH
You will have some paths separated by colons. As you will see the current path . is usually not in $PATH. So Bash cannot find your command if it is in the current directory. You can change it by having:
PATH=$PATH:.
This line adds the current directory in $PATH so you can do:
manage.py syncdb
It is not recommended as it has security issue, plus you can have weird behaviours, as . varies upon the directory you are in :)
Avoid:
PATH=.:$PATH
As you can “mask” some standard command and open the door to security breach :)
Just my two cents.
Your script, when in your home directory will not be found when the shell looks at the $PATH environment variable to find your script.
The ./ says 'look in the current directory for my script rather than looking at all the directories specified in $PATH'.
When you include the '.' you are essentially giving the "full path" to the executable bash script, so your shell does not need to check your PATH variable. Without the '.' your shell will look in your PATH variable (which you can see by running echo $PATH to see if the command you typed lives in any of the folders on your PATH. If it doesn't (as is the case with manage.py) it says it can't find the file. It is considered bad practice to include the current directory on your PATH, which is explained reasonably well here: http://www.faqs.org/faqs/unix-faq/faq/part2/section-13.html
On *nix, unlike Windows, the current directory is usually not in your $PATH variable. So the current directory is not searched when executing commands. You don't need ./ for running applications because these applications are in your $PATH; most likely they are in /bin or /usr/bin.
This question already has some awesome answers, but I wanted to add that, if your executable is on the PATH, and you get very different outputs when you run
./executable
to the ones you get if you run
executable
(let's say you run into error messages with the one and not the other), then the problem could be that you have two different versions of the executable on your machine: one on the path, and the other not.
Check this by running
which executable
and
whereis executable
It fixed my issues...I had three versions of the executable, only one of which was compiled correctly for the environment.
Rationale for the / POSIX PATH rule
The rule was mentioned at: Why do you need ./ (dot-slash) before executable or script name to run it in bash? but I would like to explain why I think that is a good design in more detail.
First, an explicit full version of the rule is:
if the path contains / (e.g. ./someprog, /bin/someprog, ./bin/someprog): CWD is used and PATH isn't
if the path does not contain / (e.g. someprog): PATH is used and CWD isn't
Now, suppose that running:
someprog
would search:
relative to CWD first
relative to PATH after
Then, if you wanted to run /bin/someprog from your distro, and you did:
someprog
it would sometimes work, but others it would fail, because you might be in a directory that contains another unrelated someprog program.
Therefore, you would soon learn that this is not reliable, and you would end up always using absolute paths when you want to use PATH, therefore defeating the purpose of PATH.
This is also why having relative paths in your PATH is a really bad idea. I'm looking at you, node_modules/bin.
Conversely, suppose that running:
./someprog
Would search:
relative to PATH first
relative to CWD after
Then, if you just downloaded a script someprog from a git repository and wanted to run it from CWD, you would never be sure that this is the actual program that would run, because maybe your distro has a:
/bin/someprog
which is in you PATH from some package you installed after drinking too much after Christmas last year.
Therefore, once again, you would be forced to always run local scripts relative to CWD with full paths to know what you are running:
"$(pwd)/someprog"
which would be extremely annoying as well.
Another rule that you might be tempted to come up with would be:
relative paths use only PATH, absolute paths only CWD
but once again this forces users to always use absolute paths for non-PATH scripts with "$(pwd)/someprog".
The / path search rule offers a simple to remember solution to the about problem:
slash: don't use PATH
no slash: only use PATH
which makes it super easy to always know what you are running, by relying on the fact that files in the current directory can be expressed either as ./somefile or somefile, and so it gives special meaning to one of them.
Sometimes, is slightly annoying that you cannot search for some/prog relative to PATH, but I don't see a saner solution to this.
When the script is not in the Path its required to do so. For more info read http://www.tldp.org/LDP/Bash-Beginners-Guide/html/sect_02_01.html
All has great answer on the question, and yes this is only applicable when running it on the current directory not unless you include the absolute path. See my samples below.
Also, the (dot-slash) made sense to me when I've the command on the child folder tmp2 (/tmp/tmp2) and it uses (double dot-slash).
SAMPLE:
[fifiip-172-31-17-12 tmp]$ ./StackO.sh
Hello Stack Overflow
[fifi#ip-172-31-17-12 tmp]$ /tmp/StackO.sh
Hello Stack Overflow
[fifi#ip-172-31-17-12 tmp]$ mkdir tmp2
[fifi#ip-172-31-17-12 tmp]$ cd tmp2/
[fifi#ip-172-31-17-12 tmp2]$ ../StackO.sh
Hello Stack Overflow

How to make my Linux C program accessible from bash

Say I made and compiled a small program in C to count the bytes of a file, called filebyte. To run it I would use ./filebyte
Now I want to make it universal on bash, like for example to run a php file, I would use bash command php file.php, same way I would like to run my program, filebyte filename.
How do I do this?
Thanks!
I often create a bin/ directory in my home directory, for small custom applications.
You then need to add that directory to your PATH, which is a list of colon-separated paths that your shell searches for executables when you type a name on thr command line.
This is usually accomplished by putting this in your ~/.bashrc file:
PATH="$PATH:~/bin"
Check the environment variable PATH and put the executable in one of the directories listed. You can also put it in a custom directory and then append it to PATH. You can check it by executing printenv PATH
If you want it for your current active shell alone, do
export PATH=$PATH:</path/to/file>
For permanently making the file available add the above line to ~/.bashrc
Why add it in PATH variable, man bash says why,
PATH The search path for commands. It is a colon-separated list of
directories in which the shell looks for commands (see COMMAND
EXECUTION below). A zero-length (null) directory name in the
value of PATH indicates the current directory. A null directory
name may appear as two adjacent colons, or as an initial or
trailing colon. The default path is system-dependent, and is set
by the administrator who installs bash. A common value is
''/usr/gnu/bin:/usr/local/bin:/usr/ucb:/bin:/usr/bin''.

How to explain './' prefix before the name of an executable in terminal? [closed]

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Every time I compile a C program, say with the cc compiler, I get an executable in the current directory. Now if I want to run it, instead of typing out just a.out or name_of_executable, I have to prefix that with this combination ./a.out.
I understand the meaning behind . (link to the current directory) and .. (link to its parent directory), also that / is a separator between directory names.
But what is the meaning of ./? Is it just a formal way to separate between referring to a name of something, and the intention to run that something (because you can't have / in your filename)?
When you type the name of an executable program, the shell searches for it in a sequence of directories whose names are stored in the $PATH environment variable.
The current directory, ., normally isn't (and shouldn't be) in your $PATH. So if you type just a.out, the shell won't find it -- unless there happens to be a /usr/bin/a.out, or /usr/local/bin/a.out, or ....
Why shouldn't . be in your $PATH? (Sometimes, on some systems, it is.) Because it creates a security hole. If . is at the front of your $PATH, then if you cd to a directory that happens to contain a command called ls, it's very easy to execute it accidentally, with arbitrarily bad consequences. Even if . is at the end of $PATH, you can still run into problems if the little program you just compiled, or the script you just wrote, happens to have the same name as a standard command. I've seen a lot of people becoming very confused because the named a test program test, but typing test runs /bin/test.
A command can be either a built-in command (provided by the shell) or the name of an executable file. If you type ls, the shell finds an executable file called ls in one of the directories named in your $PATH. If you type a.out, there is no executable file of that name in any of the directories named in $PATH -- thus the error message.
Typing a path to the executable file (even a relative path like ./a.out) causes the shell to bypass the $PATH search; you're telling the shell exactly where to find the executable file rather than asking the shell to search for it.
For security reasons, most UNIX's don't allow running a program from the current directory, only from your PATH variable. Imagine if you un-tarred a file that contained a command such as ls or gcc and it had a trojan horse installed? You could easily run it on accident.
Therefore, you have to use ./ to specify you want to run a command from the current directory.
When you enter a name of the executable, UNIX appends the name to elements of the PATH variable to obtain the path to the executable. The current directory is not searched by default. When you prefix ./ to the name, you instruct the shell to look for the executable in the current directory denoted by a dot .
Note that you could add the current directory to the PATH in order to avoid typing ./ prefix, but this is not recommended.

Executing a compiled C program

I compiled a silly little "hello world" C program called main.c:
gcc main.c
As expected, a file called a.out appeared, which they say is an executable. From that same directory, if I type
a.out
and hit enter, it says "command not found". But if I type
./a.out
It says "hello world", as desired. I've never seen an executable that requires a './' in front of it to run. Why now?
All executables that aren't in your PATH require an explicit path from root / or the local directory ./ to run. A quick search turns up other threads with essentially the same question:
Why do you need ./ (dot-slash) before script name to run it in bash?
This also has the added benefit of helping with your auto completion in your shell (assuming it supports it). If you type just aTabTab then it will list every executable in your path that starts with "a". However, if you type ./aTab it will probably just auto-complete as a.out since it will only look at executable files in the current directory starting with "a". So, looking at it that way, the "./" actually saves you typing a few keys!
It is standard practice in Unix and Linux not to have the current working directory in the path. If you want to have MSDOS/Windows behavior, alter your PATH variable to include . as the first directory.
It's because the system is looking for a.out or any other exec. file in some special paths. And the current dir in not in that list by default (usually).
look at the list of such paths:
$ env|grep PATH
you can add such current dir to PATH env. variable:
$ export PATH=$PATH:.
But you better avoid doing that and run ./a.out.
Such tech. provides us understanding that we are running specified file from current dir,
not the other file with the same name from another (potentially) dir. So, we know what we run exactly.
When you type something like a.out into a Linux terminal, you're implying that you want to run a command called a.out. By default, the terminal does not look in the current directory for these commands, it looks in PATH - a set of directories for executable programs. It is usually these directories:
/bin
/usr/bin
/usr/local/bin
among others (you can check them all by running echo $PATH)
You have to specifiy the directory directory of your program for it to run, if it is not in one of the directories of PATH. For example:
./a.out works because . refers to the directory you're in
../a.out could work if a.out is in a parent directory (.. refers to the parent)
directory
projectdir/a.out also works, if your program is in the sub-directory, projectdir
That's because a.out is not in your $PATH.
The command you provide is searched in the $PATH (environment variable in linux) by the shell.
$PATH basically is the list of directories. When you provide the executable name, shell searches it in the directories provides by $PATH.
Since a.out is not in your $PATH, you've to explicitly provide the path to a.out.

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