So I was writing a for loop and getting some errors, to get an understanding of the errors I wrote this
#! /bin/bash b=${1:- 10} echo $b for i in {0..$b} do echo "$i" done so if I run ./forloop.sh 10
I get
10 {0..10} why doesn't the range work when I have a variable as the second argument?
Bash doesn't expand the range. Use this instead.
for (( i=0; i<=$b; i++)) The part of bash that expands things like {1..10} into 1 2 3 4 5 6 7 8 9 10 runs before any parameters like $b are replaced by their values. Since {1..$b} doesn't look like a numeric range, it doesn't get expanded. By the time parameter expansion turns it into {1..10}, it's too late; nothing is going to come along and evaluate the curly-brace expression.
Change the script to use the following (http://ideone.com/MwAi16).
b=10 for i in $(eval echo {0..$b}) 
eval if you can at all avoid it. If the source of $b is even a little more remote than the immediately-prior assignment here, it's far too easy to wind up with code doing completely unexpected things.
{0..$b}, and eval is just a way to solve it. There are numerous ways to write a loop, and it is just a solution. I do not see the reason for the down vote though.
eval can possibly be a security threat au should by avoided. Also your solution can be replaced with this: b=10; for i in $(seq 0 $b); do echo $i; done. No needs for the eval. eval is evil.
for i in $(seq 0 ${1:-10}).