So I am trying to write my first shell script that can automatically run some C codes for me. I read some materials online and here is my short shell script:
#!/bin/sh
# script for grading assignment 3
echo -n "Enter the student's index >"
read index
echo "You entered: $index"
#### Functions
function question_one
{
gcc -pthread -o $index.1 $index.1.c
taskset -c 1 ./$index.1 5 5
}
#### Main
$(question_one)
As you can see, the shell script is quite simple and what it does is also quite easy to understand. First compile a C source file named like 1.1.c, 2.1.c or 3.1.c and then run the output file with just one single CPU.
When I run this script, looks like it can successfully compile the file but unable to run the output file correctly. The error message is "assignment_three_grading: line 18: Thread: command not found". However, if I type in the commands manually by myself, everything is fine.
$(question_one)
Change this to simply:
question_one
To invoke a function you just name it as if it were a regular command. Using $(...) captures its output and tries to execute that output as another command name: definitely not what you want here.
Related
I have a C program that I want to run without having to manually type commands into. I have 4 commands (5 if you count the one to exit the program) that I want given to the program and I don't know where to start. I have seen some stuff like
./a.out <<<'name'
to pass in a single string but that doesn't quite work for me.
Other issues I have that make this more difficult are that one of the commands will give an output and that output needs to be a part of a later command. If I had access to the source code I could just brute force in some loops and counters so I am trying to get a hold of it but for now I am stuck working without it. I was thinking there was a way to do this with bash scripts but I don't know what that would be.
In simple cases, bash script is a possibility: run the executable in coproc (requires version 4). A short example:
#!/bin/bash
coproc ./parrot
echo aaa >&${COPROC[1]}
read result <&${COPROC[0]}
echo $result
echo exit >&${COPROC[1]}
with parrot (a test executable):
#!/bin/bash
while [ true ]; do
read var
if [ "$var" = "exit" ]; then exit 0; fi
echo $var
done
For a more serious scenarios, use expect.
I have a very basic shell script containing
#!/bin/sh
NAME[0]="Hello"
echo ${NAME[0]}
So when I run this script an error occurs stating
./test.sh: 2: ./test.sh: NAME[0]=Hello: not found
./test.sh: 3: ./test.sh: Bad substitution
So basically I looked through a few tutorials and found this to be the basic way to declare arrays. So I am confused as to why this is an error. Any ideas?
You are starting your script as #!/bin/sh, which has a soft link to dash (The current version of sh which conforms with the POSIX 1003.2 and 1003.2a specifications for the shell) and dash doesn't support arrays. In debian 8 onwards dash has become default shell, so if you run ls -la /bin/sh the output will be /bin/sh -> dash
However bash still remains the default login shell, only the default /bin/sh used in shell scripts has been changed. So if you run your code on terminal, it will work just fine. More information on why this switch was made in Ubuntu can be found here.
If you want to use arrays in your script then you must start your script with #!/bin/bash
So your script works perfectly if modified like this
#!/bin/bash
NAME[0]="Hello"
echo ${NAME[0]}
More information on Dash as Sh DashAsBinSh
I have a script in unix that looks like this:
#!/bin/bash
gcc -osign sign.c
./sign < /usr/share/dict/words | sort | squash > out
Whenever I try to run this script it gives me an error saying that squash is not a valid command. squash is a shell script stored in the same directory as this script and looks like this:
#!/bin/bash
awk -f squash.awk
I have execute permissions set correctly but for some reason it doesn't run. Is there something else I have to do to make it able to run like shown? I am rather new to scripting so any help would be greatly appreciated!
As mentioned in #Biffen's comment, unless . is in your $PATH variable, you need to specify ./squash for the same reason you need to specify ./sign.
When parsing a bare word on the command line, bash checks all the directories listed in $PATH to see if said word is an executable file living inside any of them. Unless . is in $PATH, bash won't find squash.
To avoid this problem, you can tell bash not to go looking for squash by giving bash the complete path to it, namely ./squash.
I am working on remote computer. The shell is tcsh with no root privileges. I have no control on the server.
I have a script script.pl. When running it:
perl script.pl
gives the correct results
but when I'm trying to send this output to some program I got an error "Illegal variable name."
./vuln $(perl script.pl)
Illegal variable name.
When i'm working on gdb, the error of course repeats
(gdb) r $(perl script.pl)
Starting program: /vuln $(perl script.pl)
Illegal variable name.
How to send the output of the script to the program either on shell and gdb?
tcsh does not understand this substitution syntax $(...)
You need to use backticks for portable shell scripts:
./vuln `perl script.pl`
I have written the following code:
#!/bin/bash
#Simple array
array=(1 2 3 4 5)
echo ${array[*]}
And I am getting error:
array.sh: 3: array.sh: Syntax error: "(" unexpected
From what I came to know from Google, that this might be due to the fact that Ubuntu is now not taking "#!/bin/bash" by default... but then again I added the line but the error is still coming.
Also I have tried by executing bash array.sh but no luck! It prints blank.
My Ubuntu version is: Ubuntu 14.04
Given that script:
#!/bin/bash
#Simple array
array=(1 2 3 4 5)
echo ${array[*]}
and assuming:
It's in a file in your current directory named array.sh;
You've done chmod +x array.sh;
You have a sufficiently new version of bash installed in /bin/bash (you report that you have 4.3.8, which is certainly new enough); and
You execute it correctly
then that should work without any problem.
If you execute the script by typing
./array.sh
the system will pay attention to the #!/bin/bash line and execute the script using /bin/bash.
If you execute it by typing something like:
sh ./array.sh
then it will execute it using /bin/sh. On Ubuntu, /bin/sh is typically a symbolic link to /bin/dash, a Bourne-like shell that doesn't support arrays. That will give you exactly the error message that you report.
The shell used to execute a script is not affected by which shell you're currently using or by which shell is configured as your login shell in /etc/passwd or equivalent (unless you use the source or . command).
In your own answer, you say you fixed the problem by using chsh to change your default login shell to /bin/bash. That by itself should not have any effect. (And /bin/bash is the default login shell on Ubuntu anyway; had you changed it to something else previously?)
What must have happened is that you changed the command you use from sh ./array.sh to ./array.sh without realizing it.
Try running sh ./array.sh and see if you get the same error.
Instead of using sh to run the script,
try the following command:
bash ./array.sh
I solved the problem miraculously. In order to solve the issue, I found a link where it was described to be gone by using the following code. After executing them, the issue got resolved.
chsh -s /bin/bash adhikarisubir
grep ^adhikarisubir /etc/passwd
FYI, "adhikarisubir" is my username.
After executing these commands, bash array.sh produced the desired result.