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Returning Strings from C program to Unix shell script

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# 1  
MySQL Returning Strings from C program to Unix shell script


I'm having a requirement where I need to call a C program from a shell script and return the value from the C program to shell script.
I refered a thread in this forum. But using that command in the code, it is throwing an error


Error: bash: get_password : command not found

I have compiled the code get_password.c , the executable is generated and there's only one display statement in the code.

And also please let me know whether I can pass any arguments to the C program or not. If so, what's the way to do that.

Thanks for the help in advance.


Last edited by venkatesh_sasi; 11-24-2007 at 05:00 AM.. Reason: subject line change
# 2  
is the binary get_passwd available in the expected directory / the directory from which the shell script runs ?

if not, make sure to have the absolute path of the binary in the shell script

Yes, you can pass arguments to binary ( C program )

./run arg1 arg2 arg3

catch in the program with argv
# 3  
BTW shouldn't this be a new thread ?
# 4  
Originally Posted by matrixmadhan
BTW shouldn't this be a new thread ?
Right you are... split and moved to a new thread.
# 5  
The shell doesn't know where to find your command "get_password".
You could put it into a well known location (eg. "/usr/local/bin", but beware that that directory is often insecure so someone might be able to substitute a trojan in there), or add the current directory (".") to the PATH (which is a very bad thing to do!).
My preference is to create a new specific "bin" directory for bespoke tools (we use "/opt/sup/bin" in my current place of work), and add it to the PATH.
# 6  
Originally Posted by venkatesh_sasi
I would expect...


# 7  
Hi All,

Thanks for the replies.
Please see the below shell script.
# ! /bin/bash
echo " In the program"
set output = '/root/Desktop/get_password'
if [ $? -eq 0 ]
echo $output
output: In the program

After displaying the statment "In the program",
it's just displaying a blank line.
I tried putting the statment echo ${output} also, But it's of no use.
Can I give
set output ='/root/Desktop/get_password arg1 arg2'
where arg1 and arg2 are the mandatory arguments that need to be passed to the get_password.

Note : The executable file get_password is there at the above mentioned location and also I'm printing only the password at the console.

Thanks & regards,

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