UNIX for Advanced & Expert Users

Say there is a shell script named test.sh. I intentionally omit the #! line in test.sh for testing perpose. I did the following :

$ echo $0
-ksh ---> current shell
$ echo $$
12919 ---> PID of the current shell
$./test.sh
----> test.sh contains a sleep command to prevent it from terminating immediately

Then I switched to another terminal session to continue my testing:

Code:

$ ps -ef|grep 12919|grep -v grep
wesadmin  5083 12919  0 10:10:27 pts/3     0:00 -ksh
wesadmin 12919  7616  0 08:10:37 pts/3     0:00 -ksh

You can see that the process id associated with the script test.sh is 5083.

Questions here:

1. Why shell itself displayed in the process list without test.sh as the argument? We usually expect " -ksh test.sh " displayed. My another testing shows that the argument test.sh can be displayed if #! line added in the script.

2. Why the invoked shell is ksh instead of ssh? To my knowledge,, if #! line not specified explicitly, the default shell, i.e. sh is used as the interpreter .


Thanks!

My guess would be that without the shebang line (#!), the script automatically is executed by the current shell (maybe in a subshell as it seems from the example you gave). With the shebang line the script is passed to an exec call, where the system, instead of the currently running shell, decides which interpreter to start.

See: https://www.unix.com/tips-tutorials/3...r-bin-ksh.html

Yes, the following words from The Whole Story on #! /usr/bin/ksh can explain my doubt ( in fact, I also guessed ksh works this way for this case)

However my confusion is still not completly clarified. I aways told that if #! line is not specified, sh is invoked as the default interpreter. Is it wrong? And for the statment '...ksh will try to exec the script and fail ...', why exec fail? I did some test by making a small c program in which I try to execute a script without #! line using execl ( ...) system call, it works! In other words, execl can execute this kind of script successfully and morever the sh is invoked as defaut shell as the interpreter of the script! Then why doesn't ksh use exec() system call directly to handle a script without #! line? And also, I did adittional tests in posix-sh and bash enviroment, seems that both of them treat a script without shabang in the same way as ksh. Again, why don't they exec() directrly to execute the script?

Your remaining questions have mostly been answered in the linked article. But as for why things are this way... A shell script cannot truly be exec'ed. A program must be in machine language for the kernel to be able to transfer control to it. The machine language is very close to assembler. A machine language program will just be a binary file. The cpu cannot directly run a shell script. The kernel must actually exec an interpreter (some shell) to run the shell script.