arrays - Fill an array with a loop in BASH

Question

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?

Answer 1

您可以使用+=运算符：

for i in $(seq 200 5 4800); do
    array+=($i)
done

Answer 2

您可以简单地：

array=( $( seq 200 5 4800 ) )

你已经准备好你的阵列了。

Answer 3

用bash的方式来做：

array=( {200..4800..5} )

Answer 4

这些方法可能会导致内存（或行的最大长度）问题，所以这里有另一个：

# 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 
}

像这样使用：

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