Suppose that I add a program to path that is dependent on a file name "test.in". I programmed this in C++ so I used ifstream fin("test.in") without specifying the directory. Now if I were to run this program from a different directory, would the program be able to access the file "test.in"?
Firstly, this has nothing to do with the file extension, which is merely a convention given as part of the filename.
Secondly, you were always using a relative path. Even when you were running your program "from the same directory" as test.in, you were reliant on the "working directory" of your shell context being the same as the directory in which the executable and the file reside.
This is not always the case.
For example:
~/myProject:# ls
test.in
program
~/myProject:# ./program
This is okay, because your shell is at ~/myProject, and so is test.in.
However, if you'd written:
~/myProject:# cd ..
~:# ./myProject/program
…then your test.in file wouldn't be found, as it does not exist in ~. It exists in ~/myProject. It doesn't matter that the executable itself is also found in ~/myProject.
This is actually desirable behaviour, as it allows flexibility from the shell. Ideally you would allow support for piping/redirecting the file to the process instead (program < test.in — now there are no assumptions baked into your code at all!), but we can save that for another day.
For now, you seem to be concerned about what happens if you move the executable away. Don't worry: just use this feature!
~:# mv myProject/program .
~:# cd myProject
~/myProject:# ../myProject
Your working directory is the directory in which test.in resides, so it will be found via the relative path given in your program code.
Related
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
I have a c++ project that I would like to send to someone in executable form. The issue is the program must read from a .txt that I created (specific deliminators). Currently my program reads from a file path that is specific to my computer,
parseFile("/Users/David/Desktop/FinalProject/Store.txt");
How could I package the .txt file and the executable file together, where the exec. reads specifically from the that .txt on anyone's machine?
Note: I am using Xcode
Change your programs to receive 'file path' as a parameter. Write a note(ReadMe) with the program to specify the file format and added a sample data file with the package
tl;dr: if you just put the text file in the same folder with your executable, you can open it with parseFile("Store.txt");
In most runtime implementations, there is a notion of a "working directory." When you open up an executable via the graphical shell (by double clicking it or something to that effect) the working directory is the same as the directory the executable is in.
Now, if you try to open a file in your program via a path that isn't fully qualified, then the path that gets used will be relative to the working directory.
A fully qualified path is a discrete path that points to a single entity in your filesystem. "/Users/David/Desktop/FinalProject/Store.txt" is one such example, as it starts at root (/ on *nix, DriveLetter:\ on Windows) and says exactly which directories you need to traverse to get to your file.
A path that is not fully qualified (which basically means that it doesn't start at the root of your filesystem) can be used to perform relative file addressing. Most runtimes will assume that any path that is not fully qualified is meant to be relative to the working directory, which basically means that the path that actually gets opened is the result of concatenating your provided path to the end of the working directory.
As an example, if you opened your binary, which is stored as /Users/David/Desktop/FinalProject/a.exe, then the working directory would be set to /Users/David/Desktop/FinalProject/. If your program then tried to open "Store.txt", the runtime would see that you're trying to open a path that isn't fully qualified, so it would assume you meant to open a file relative to the working directory, which would then be /Users/David/Desktop/FinalProject/ + Store.txt, which would be /Users/David/Desktop/FinalProject/Store.txt.
The nice thing about this is that if you move your binary, the working directory moves too. if you move a.exe along with Store.txt to /Users/David/Desktop/FinalProject(copy)/, then when you open /Users/David/Desktop/FinalProject(copy)/a.exe, the working directory will be /Users/David/Desktop/FinalProject(copy)/ now, and now when you call parseFile("Store.txt"), it will instead open up /Users/David/Desktop/FinalProject(copy)/Store.txt. This holds true when moving to other computers, too.
It's worth noting that if your binary is run from a command line utility, the working directory will often be the directory the command line shell is in, rather than the executable's directory. It is, however, a part of the C standard that the first command line parameter to main() should be the name of the executable, and most implementations supply you with the fully qualified path. With some minimal parsing, you can use that to determine what path to use as a base for addressing files.
I'm quite confused about this weird behaviour of my .cpp project. I've got the following folder structure:
include/mylib.h
myproject/src/eval.cpp
myproject/data/file.csv
myproject/Makefile
In eval.cpp I include mylib.h as follows:
#include "../../include/mylib.h"
and compile it through Makefile:
all:
g++ -I include ../include/mylib.h src/eval.cpp -o eval.out
Now in my eval.cpp I'm reading the file.csv from data directory and if I refer to it like this
../data/file.csv
it doesn't find it (gets empty lines all the time), but this
data/file.csv
works fine.
So, to include mylib.h it goes two directories up (from src folder) which seems right. But it doesn't make sense to me that to refer to another file from the same piece of code it assumes we are in project directory. I suppose it is connected with Makefile somehow, but I'm not sure.
Why is it so?
EDIT: After a few thing I tried it seems that the path which is used is not the path from binary location to the data location, but depends on where from I run the binary as well. I.e., if I have binary in bin directory and run it like:
./bin/eval.out
It works with data/file.csv.
This:
cd bin
./eval.out
works with ../data/file.csv.
Now it seems very confusing to me as depending on where I run the program from it will give different output. Can anyone please elaborate on the reasons for this behaviour and if it is normal or I'm making some mistake?
It is so because (as explained here ) the compiler will search for #included files with quotes (not with brackets) with the current working directory being the location of the source file.
Then, when you try to open your .csv file, it's now your program that looks for a file. But your program runs with the current working directory being myproject/ which explains why you must specify data/file.csv as your file path, and not ../data/file.csv. Your program does not run in your src folder, it will run in the directory the binary ends up being invoked from.
You could have noticed that in your Makefile, your -I options specify a different path for your header file than your .cpp file.
EDIT Answer: It's quite simple actually and completely normal. When you invoke your binary, the directory which you're in is the current working directory. That is, if you run it with the command ./myproject/bin/eval.out, the current working directory is . (e.g. /home/the_user/cpp_projects). My post was a bit misleading about that, I corrected it.
Note: You can use the command pwd in a command prompt to know which is the current working directory of this prompt (pwd stands for "print working directory").
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Possible Duplicate:
What should Linux/Unix 'make install' consist of?
I'm making a program that can be invoked from the command line, like ./prog arg1 arg2. I was wondering, how can I make it so that I can run it from anywhere on the system? I know that I could put prog into /usr/bin/, but what if my program needs resources from its install directory (that can be wherever the user downloaded it)?
put the directory in which your program resides into the path environment variable or move your program into one of the directories already in path (usually requires superuser permission, which I gather you don't have for then you wouldn't ask this question).
to add a directory to the front of the search path and have the system refresh its database on tcsh, say
setenv "my/directory:"$PATH
rehash
on bash, I think, it's
PATH=/my/directory:$PATH
export PATH
(no need to rehash). Note that the above commands put your directory at the top of the search path, i.e. these will be searched before any other. Thus, if your program is called "gcc", then your program will be executed rather than the GNU C compiler. Alternatively, you can add your directory to the end of the search path, in which case your program will only be picked up if no other program of the same name is found in any of the other directories in the search path.
You probably also want to become familiar with the Linux Filesystem Hierarchy: the standard definition for "what goes where". Here's more information:
https://superuser.com/questions/90479/what-is-the-conventional-install-location-for-applications-in-linux
Environment variables can be defined globally ("for everybody", e.g. /etc/profile), or locally ("per user", e.g. ~/.bashrc). Here's a good summary of some of your options:
https://wiki.archlinux.org/index.php/Environment_Variables
When you execute a programme using prog arg1 arg2, it's thanks to your shell, which search in the $PATH environement variable for folders where programs are. (Try env | grep PATH to see those folder).
You need eather to add a new directory in this variable (export PATH="/new/directory/path/:$PATH" if under bash, setenv PATH "/new/directory/path/:$PATH" if with tcsh) or copy your program and all the files it need to execute in one of the PATH folder.
There are two ways of dealing with this (and Makefiles have nothing to do with them)
Your installer could just put the files where it wants them, so your program doesn't have to search -- it can use hardcoded paths. Or you could put the path to the data directory into yet another file, which would be hardcoded (like /etc/programname.config).
You put all your stuff into one directory (often something like /opt/programname). You can hardcode that too, of course, or your program can readlink() the /proc/pid/exe file for a good chance (no guarantee, though. In particular, it works if for example a symlink is used to point from /usr/bin/programname to your /opt/programname/bin/programname or whatever, but it won't work if that's a hardlink)
to get the path to the executable. From there you should be able to reach your data files.
If prefer the second solution, but that's just me. The first solution works well with package managers, and it's less overkill if you don't really have a lot of data files.
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If you look at point (6) here: http://www2.warwick.ac.uk/fac/sci/moac/students/peter_cock/cygwin/part3/
Why should we type ./ before the .exe file in order for it to run?
Why cannot we type hello.exe immediately?
Thanks.
Usually because intelligent people don't have their current directory . on the path :-)
The path is an environment variable like /bin:/usr/bin:/usr/sbin, and it's a list of directories to look in for finding executables, such as when you type in hello.
Unlike Windows, many UNIX shells don't automatically search the current directory for an executable. They must be listed in the path otherwise they are not run.
That's because to do otherwise is actually an attack vector. For example, if you create an ls program in your home directory and tell one of the administrators that there's a funny file in there, they may go to your directory and enter ls to see what's in there.
For a silly administrator that has the current directory before the "real" location of ls, they are now compromised, because your code is running with their full privileges.
That's why they tend not to do that.
Some people (not I) will put . on their path to make their lives easier but, even then, they'll put it at the end so that other locations are searched first.
Administrators don't have the luxury of being that trusting.
Because the current working directory is not in the PATH?
Or at least, that's how things are setup on Unix-style systems, I assume CYGWIN does the same.
On Windows, the current directory is always in the search path for an executable. The search order is "look in the current dir, if not found, look in the directories listed in the PATH environment variable".
From MS site:
The operating system always searches
in the current directory first, before
it searches the directories in the
command path.
(which makes all the warning here of not putting the . in your PATH irrelevant, IMHO)
On Linux this is not the case (for current dir). So, to run an executable which is in your current dir you need to write ./exe_name.
As Cygwin, again AFAIK, is for Windows, the ./ is not needed and seems to be just a copy/paste or preserving the unix-style the writer is used to.
EDIT: this is the issue of the command processor (the shell) as pointed out in comments and as I explain below, so if you are using a Unix-like shell on Windows, you still may need this style.
EDIT: elaborating on .\
. (not ./ to be exact) is an alias to the current directory. On Unix, every newly created directory is not "born" empty but contains 2 children: ., which is a self-reference, and .. which is a reference to the parent directory. Both are just regular directories, as any other. You don't see them when you run the ls command (same as dir on Windows) because names starting with . are special in the sense that they are not displayed by default. However, you can see them by ls -a.
When you run a command at the prompt, if the command is only a (file) name, the system (actually, the shell) searches the PATH for the file with this name.
If the command contains a path (not necessarily an absolute path, e.g. subdir1/exe) the system looks for the executable where you specified. Hence, writing ./exe means file exe in the current dir.
Cygwin is a Unix-like runtime environment and as such follows the way paths are searched for executables in such environments. The default executable search path of Unices does not contain the current directory. Thus if one wants to run an executable not located in one of the directories set in PATH a full path must be given. ./ is a shorthand for the current directory, also called process working directory (pwd). Be advised that it's a very bad idea to have the pwd being included in the executable search path.
Cygwin follows the Unix limitations on executing files in the current working directory. In Unix style terminal environments an executable must have ./ prepended if it is to be executed from the current directory. This is because the current directory "." is not part of the PATH environmment in order to limit the damage done by malware. Cygwin is simply following this convention, it has nothing per say to do with C++ programs
That's just an issue with your 'path' or 'PATH' variable in your shell. (probably your shell is bash, so it'd be PATH.)
echo $PATH
A typical 'user' path to do what you want would start with "." as a path element. This is a minor security risk of course.