I would like to fill an array automatically in bash like this one:
200 205 210 215 220 225 ... 4800
I tried with for like this:
for i in $(seq 200 5 4800);do
array[$i-200]=$i;
done
Can you please help me?
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4 Answers
You can use += operator:
for i in $(seq 200 5 4800); do
array+=($i)
done
Comments
Do it the bash way:
array=( {200..4800..5} )
answered Jun 28, 2013 at 11:13
gniourf_gniourf
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Comments
You could have memory (or maximum length for a line) problems with those approaches, so here is another one:
# function that returns the value of the "array"
value () { # returns values of the virtual array for each index passed in parameter
#you could add checks for non-integer, negative, etc
while [ "$#" -gt 0 ]
do
#you could add checks for non-integer, negative, etc
printf "$(( ($1 - 1) * 5 + 200 ))"
shift
[ "$#" -gt 0 ] && printf " "
done
}
Used like this:
the_prompt$ echo "5th value is : $( value 5 )"
5th value is : 220
the_prompt$ echo "6th and 9th values are : $( value 6 9 )"
6th and 9th values are : 225 240
