Bash script for loop which the last statement is first calculated

Question

I wrote this bash script

x=64
y=1
ans=$((x-y))
z=`expr $ans`
for i in {1..$z}
do 
    echo $i
done

Actually, I would like to print i values from 1 to 63, 63 which is first obtained from the above addition. But it just prints {1..63}

Can someone please help me. Thanks in advance.

**Please don't answer this question. This is a 1:1 duplicate** of [How do I iterate over a range of numbers defined by variables in Bash?](https://stackoverflow.com/questions/169511/how-do-i-iterate-over-a-range-of-numbers-defined-by-variables-in-bash). Which has been asked over and over again. — Socowi, Mar 13 '19 at 10:30
What are you trying to do with ``z=`expr $ans` ``? `z=$((x-y))` or just `((z=x-y))` is sufficient. — Socowi, Mar 13 '19 at 10:32

score 0 · Answer 1 · answered Mar 13 '19 at 10:30

0

You're best of using a C-style for loop:

for ((n=$y; n<$x; n++)); do
    echo $n
done

answered Mar 13 '19 at 10:30

Robert Seaman

score 0 · Answer 2 · answered Mar 13 '19 at 10:33

0

using seq :

for i in $(seq 1 $z) 
do 
   echo $i 
done

answered Mar 13 '19 at 10:33

nullPointer

2 Answers2