Question

How to create an array with a sequence of numbers in Bash

Answer 1

# seq FIRST STEP LAST

arr=($(seq 0 1 9))

for i in ${!arr[@]}; do
    echo "element $i is ${arr[$i]}"
done
 
 
 
# run:
#
# element 0 is 0
# element 1 is 1
# element 2 is 2
# element 3 is 3
# element 4 is 4
# element 5 is 5
# element 6 is 6
# element 7 is 7
# element 8 is 8
# element 9 is 9
#

 

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Answer 2

# seq FIRST STEP LAST

arr=($(seq 0 0.2 2.4))

for i in ${!arr[@]}; do
    echo "element $i is ${arr[$i]}"
done
 
 
 
# run:
#
# element 0 is 0.0
# element 1 is 0.2
# element 2 is 0.4
# element 3 is 0.6
# element 4 is 0.8
# element 5 is 1.0
# element 6 is 1.2
# element 7 is 1.4
# element 8 is 1.6
# element 9 is 1.8
# element 10 is 2.0
# element 11 is 2.2
# element 12 is 2.4
#

 

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