Assigning a variable a filename in bash

I am writing a script to generate an executable (arm executable) in Linux by taking in user-specified .s file. So the user enters an input file, say "input.s" and an output file name, say "output.axf" and the script generates the desired output (executable - .axf). Now I want an additional option wherein, if the user does not give an output filename in the arguments, I want to create a default output file myself. The script is as follows:

#!/bin/bash echo Enter the names of the input file and output file read input_file output_file if [ -z "$input_file" ] then echo "No input supplied" elif [ -z "$output_file" ] then $output_file=brot.axf elif [ -z "$input_file" && -z "$output_file" ] then echo "No input/output file supplied" fi ifilename=$(basename "$input_file") ifilename="${input_file%.*}" armasm -g --cpu=8-A.64 "$input_file" armlink "$ifilename.o" -o "$output_file" fromelf --test -c $output_file > disassembly.txt 

Now my problem is that every time I run the script and do not specify anything for $output_file, I get this error:

./script_test.sh: line 12: =brot.axf: command not found

Fatal error: L3901U: Missing argument for option 'o'.

However, when I do specify the input and output file names with extensions, it works as expected. How do I fix the error and assign a default name to the output file if the user doesn't assign one ?

2 Answers

Variable assignments don't take the $ notation in bash shell. You just need below without the $

output_file="brot.axf" 

And later in the script if filename is a variable and trying to construct a name with .o appended, enclose the variable name within {} , so that the variable is expanded properly

armlink "${filename}.o" -o "$output_file" 

Also by the looks of it you have a likely typo in the filename as variable ifilename. If you care trying to use it double-quote it as above.

You have to remove the $ in front of output_file on line 12.

output_file=brot.axf