Unix & Linux Stack Exchange is a question and answer site for users of Linux, FreeBSD and other Un*x-like operating systems. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

with Bash's source it is possible to execute a script without execution bit set. This is documented and expected behavoir, but isn't this against the use of an execution bit?

I know, that source doesn't create a subshell.

share|improve this question
1  
The fact that chmod can let you set permissions (including `x) with an octal number gives some clue to what era it comes from. I wouldn't be surprised if it started as a quick and dirty "this is a binary file you can execute" indicator, from the days before she-bang was invented, but I have no evidence to that – infixed 2 days ago
5  
Then again, when SUID comes into play, different levels of protection for different categories of users become more important. Go ahead and read my SUID program if you want. But if you execute it via simply reading it, the SUID powers are not coming along with it – infixed 2 days ago
    
@infixed The Linux program-loader won't even look at the shebang unless the execution bit is set. (To toot my own horn a bit: see here.) – Kyle Strand 7 hours ago
up vote 9 down vote accepted

Bash is an interpreter; it accepts input and does whatever it wants to. It doesn't need to heed the executable bit. In fact, Bash is portable, and can run on operating systems and filesystems that don't have any concept of an executable bit.

What does care about the executable bit is the operating system kernel. When the Linux kernel performs an exec, for example, it checks that the filesystem is not mounted with a noexec option, it checks the executable bit of the program file, and enforces any requirements imposed by security modules (such as SELinux or AppArmor).

Note that the executable bit is a rather discretionary kind of control. On a Linux x86-64 system, for example, you can bypass the kernel's verification of the executable bit by explicitly invoking /lib/x86_64-linux-gnu/ld-linux-x86-64.so.2 as the interpreter:

cp /bin/ls /tmp/
chmod -x /tmp/ls
/lib/x86_64-linux-gnu/ld-linux-x86-64.so.2 /tmp/ls

This is somewhat analogous to sourcing Bash source code in Bash, except that ld.so is the interpreter, and the code that it executes is machine code in ELF format.

share|improve this answer
2  
Using the loader as an "interpreter" of "bytecode" that just happens to be a real executable binary.....awesome. Thank you for that. – Kyle Strand 7 hours ago
    
@KyleStrand there may be several levels of indirection (the difference between the stored program and the loaded process and how the memory is mapped, various supervisors, VMs, etc) but in principle the machine code can be executed by CPU directly (no microcode, etc) and therefore the machine code is not interpreted: the hardware understands it as is—it is the native language—no interpreters are necessary. – J.F. Sebastian 3 hours ago
1  
@J.F.Sebastian Yes, we know that it is native code, hence "interpreter" is in quotes. The job of ld.so is to do dynamic linking. – 200_success 3 hours ago
    
@J.F.Sebastian What 200_success said. Also, that's what I meant by "real executable binary," and I also put "bytecode" in quotes (since AFAIK "bytecode" is typically not used to refer to actual compiled natively-executable binary code). Also, nice username. – Kyle Strand 3 hours ago

source or the equivalent but standard dot . do not execute the script, but read the commands from script file, then execute them, line by line, in current shell environment.

There's nothing against the use of execution bit, because the shell only need read permission to read the content of file.

The execution bit is only required when you run the script. Here the shell will fork() new process then using execve() function to create new process image from the script, which is required to be regular, executable file.

share|improve this answer
1  
@Motte001: Yes, it's work with any command, just like you type in terminal for example – cuonglm 2 days ago
4  
@alexis: I'm not sure I really follow. If you do bash < script, you get essentially the same result as source script. What protection does an execute bit check provide? – Conspicuous Compiler 2 days ago
25  
@alexis, it's not an interpreter's job to check permissions of the scripts it interprets. Nothing does this -- not Python, not Ruby, not the Java virtual machine, not other random shells. The execute bit controls whether the OS execv* family of syscalls can be used with an executable, not whether an interpreter will run it. Why confuse people (and break ability to evaluate code streamed from non-file sources) by breaking conventions? – Charles Duffy 2 days ago
11  
@alexis, ...moreover, there's a valuable meaning having a shell library that's not executable: It says "I'm a library, not a command -- source me, don't execute me". Otherwise you'd need to write code to detect and remediate misuse of code intended to only be sourced, but by not having the +x permission you can ensure that users can't (won't) misuse a library in such a way. – Charles Duffy 2 days ago
4  
@alexis "it was a decision by the authors" Only in the sense that every other possible bit of code that wasn't included was a decision. The shell opens the file for reading, to read the commands out of it. I would not expect it to check for read permission, it just tries to open the file and fails if it can't. Likewise it doesn't check for write or execute permission, since those checks are not relevant for reading the file. – Randy Orrison 2 days ago

The executable bit (unlike the rest) on nonsetuid and nonsetguid files isn't much of a security mechanism. Anything you can read, you can run indirectly, and Linux will let you indirectly read anything you can run but not directly read (that should be enough to punch a hole in the concept of non-set(g)uid x-bit being a security measure).

It's more of a convenience thing: Let the system run it for me directly if the the bit is set, otherwise I need to do it indirectly (bash the_script; or some hacking to get the memory image of a no-read-permission executable).

You can set it for convenience if you intend to both in-source and execute your insourcable.

Apparently, however, many shared library implementers share your thinking and consequently, many systems do require that shared libraries, which are essentially the native equivalent of shell insourcables, be marked executable in order to be usable. See Why are shared libraries executable?.

share|improve this answer
    
Of course I can run everything when I can read it, but the execution bit can prevent "normal" dektop users from infection their PC by clicking on mail attachments (If such users would use Linux). So it is kind of a security feature – Motte001 2 days ago
3  
@Motte001 They can execute those attachments if they really want to. It's a convenience. – PSkocik 2 days ago
2  
The executable bit is not for security: If I own a file I am free to chmod it and make it executable. It is for sorting data from programs, so the OP's question is a reasonable one. – alexis 2 days ago
2  
One reason we need the executable bit is to run setuid programs, particularly those owned by root. While you can certainly execute them on your own, you need the OS to run them so that they will have the necessary permissions. In some other cases it really is just a convenience. – Waleed Khan 2 days ago
2  
"Linux will let you read anything you can run via /proc" -- Well, my stock Debian kernel doesn't: /tmp$ cp /bin/cat ./cat ; chmod a-rw ./cat ; ./cat & cp /proc/$!/exe /tmp/cat2 -> cp: cannot stat ‘/proc/16260/exe’: Permission denied – ilkkachu 2 days ago

That's a good question! Unix uses the executable bit to distinguish between programs and data. The OS does not require the execution bit, since a sourced script is not passed to the OS for execution as a new process. But the shell treats a sourced script as a program, and will look in $PATH for the file you want to source. So, the shell itself could have required execute permission for sourced files; but it didn't.

The question must have come up long ago. The design of the Bourne shell was the result of "a long sequence of modification, dialog, discussion" among the denizens of Bell Labs, and many design decisions were discussed by S.R. Bourne and others over the years. Unfortunately, my quick look didn't find any discussion of the source feature (in my defense, it's hard to google for it). What I did find is that the "." command does not appear in this early introduction to the shell by Bourne himself, but it is present in the more mature Version 7 version.

Absent authority, here's my own interpretation:

  1. The . command, aka source, is in effect textual inclusion (like #include in the C preprocessor) into the source code of the executing script or interactive session. As such, the included file is arguably not "executed".

  2. The Unix philosophy has always been to give the programmers enough rope to hang themselves. Too much hand-holding and arbitrary restrictions just get in the way. (It is only rather recently that some distributions made rm -r / refuse to do what you ask. Do not try this as root!) So, perhaps Bourne at al. just decided that when you try to source a file, you are assumed to know what you are doing. That also avoids unnecessary work, and cycles mattered a lot back then.

share|improve this answer
    
But one should definitely try rm -rf --no-preserve-root / as root! /s – cat yesterday

The distinction is important because you may have a file of shell commands which is not useful as an executable, but only useful when sourced. For this file you can turn off the execute bit and then it will never be accessed unless explicitly in a source command. The reason for such a thing is to have side effects on the shell it is run from. For a specific example, I have a script called fix_path which looks at and modifies the path.

share|improve this answer

As far as the OS is concerned, a file containing shell script is just data. If you pass the name of such a data file to the source command or pass it on the command line to an invocation of the bash shell, all the OS sees is a string that happens to coincide with the name of a file containing data.

How would the execute bit be at all relevant in that case?

share|improve this answer

Another point of view:

Sourced script basically consists of shell builtins and program calls. Shell builtins (with source among them) are parts of the shell and the shell must be executable in the first place. Every program called (that being ELF, another script with shebang, whatever) has to have execution bit set, otherwise it will not run.

So it is not against the use of an execution bit, because nothing without execution bit will run. The validation doesn't occur for the sourced script as a whole; it is performed for every part separately, but it is.

share|improve this answer
    
Shell buildins are not sourced from a shell script. Typically they are part of the shell executable. In ksh93, zsh and a couple of other shells they can be shell extensions. – fpmurphy1 yesterday
    
@fpmurphy1 Isn't this my sentence "shell builtins (...) are parts of the shell"? – Kamil Maciorowski yesterday

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.