Any ideas why this is happening? Why do I have to manually explicitly reassign the variable but can't do it if I have another variable in the name of the variable?
SCRIPT:
#!/bin/bash
a_1=1
a_2=1
for temp in 1 2
do
a_$temp="2"
echo $((a_$temp))
done
a_1=2
a_2=2
echo $a_1
echo $a_2
OUTPUT:
[dgupta@della4 Rates_Of_Quenching]$ ./test.sh
./test.sh: line 8: a_1=2: command not found
1
./test.sh: line 8: a_2=2: command not found
1
2
2
Instead of:
a_$temp="2"
Use:
declare a_$temp="2"
to create variable with dynamic name.
As far as bash is concerned, you are trying to execute the command 'a_1=2', rather than perform an assignment. You can get around this by using declare, or its synonym typeset:
'a_1=2' # bash: a_1=2: command not found
typeset 'a_1=2'
echo $a_1 # 2
declare 'a_1=3'
echo $a_1 # 3
While it is possible to use declare, you might want to take advantage of bash arrays (which have been around since bash version 2) rather than using variables with numerical suffixes:
a=(1 1)
echo ${a[0]} # 1
echo ${a[1]} # 1
for i in 0 1; do a[i]=2; done
echo ${a[0]} # 2
echo ${a[1]} # 2
bash considers typeset obsolete (it still has its place if maximum portability is required) and (b) your inclusion of a prompt string (bash-4.3$) both creates visual noise and makes it harder to copy-paste your code for verification (it's commendable to state the bash version your code was run on, but that could be done separately, although, depending on the feature discussed, stating the version may not be necessary at all, if a feature was introduced a loooong time ago).$), as it makes copy-pasting any sequence of commands cumbersome. I know that $ is seductive, because it is widely used to symbolize a prompt, whereas I know of no equally well-known symbol to indicate output (is there one?). However, I think on SO reversing the logic is beneficial: no prefix for the commands, and some indicator for output. One - not very elegant - way around this is to place output in a separate code snippet.