⟩ What is the output of this program? #!/bin/bash var1=10 $var1=20 echo $var1 exit 0 a) program will print 10 b) program will generate a warning message c) program will print 20 d) both (a) and (b)
d) both (a) and (b)
Explanation:
The doller sign ($) is used to access a variable's value, not to define it.
Output:
root@ubuntu:/home/google# ./test.sh
./test.sh: line 3: 10=20: command not found
10
root@ubuntu:/home/google#