Passing command line arguments in to a bash script command

Question

Lets say I have a variable storing the name of a compiled C++ program.
executable="myprog.out"

My script tt needs to create a transcript of the C++ program and exit it (without the user having to type exit to exit the bash script command).

So far I've achieved this by doing by calling the bash script command inside tt script like so,

#!/bin/bash
script -q -c $executable transcript

This method works provided that the C++ program doesn't take command line arguments. If I have a variable holding the command line arguments needed for the program as,
progArgs="arg1"

How do I call the script command and have it enter the command line arguments for $executable?

redneb · Accepted Answer · 2016-09-16 01:18:29Z

3

The -c option actually takes a shell command as its argument, so you can pass arguments like that:

script -q -c "$executable --opt1 --opt2 -f test" transcript

answered Sep 16, 2016 at 1:18

redneb

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2 Comments

syntax error Over a year ago

Didn't know you could do the the -- for options. Thank you!!

redneb Over a year ago

@syntaxerror Whether you use -- or - depends on the program, this was just an example.

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