What is the output of this program?
#!/bin/bash
var1=10
$var1=20
echo $var1
exit 0
a) program will print 10
b) program will generate a warning message
c) program will print 20
d) both (a) and (b)
Submitted by: Murtazad) both (a) and (b)
Explanation:
The doller sign ($) 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#
Submitted by: Murtaza
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#
Submitted by: Murtaza
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