Converting from for loop to while if fi loop UNIX - bash

Question

I'm trying to figure out how to convert the following for loop into a while loop or until loop including if fi for UNIX Here's the code...
#!/bin/bash
if [ $# -gt 0 ] ; then
start=$1
end=$3
step=$2
for x in `seq $start $step $end` ; do
number $x
done
else
echo Enter at least one number on the command line.
fi
Any help is greatly appreciated!

Answer 1

(( x = start ))
until (( x > end ))
do
number $x
(( x += step ))
done

Answer 2

For a loop like this, it's easiest to use the numeric-style for loop:
for ((x=start; x<=end; x+=step)); do
number $x
done

Answer 3

Using seq is good as it has some default values for step and end value. I added this to the if if only one, or two arguments are given. Also it handles the situation if the given argument are not numbers (default values are 0). Also an endless loop may encounter if step is zero. If You remove this checking it will work the same way as the seq version.
if [ $# -gt 0 ] ; then
x=$(($1+0))
end=$((${3:-$x}+0))
step=$((${2:-1}+0))
[ $step -eq 0 ] && echo "Step is zero!" && exit 1
while [ $x -le $end -a $step -gt 0 -o $x -ge $end -a $step -lt 0 ]; do
echo $x
x=$((x+step))
done
else
echo Enter at least one number on the command line.
fi
Change echo to number in the while loop if you would like to use!
Examples:
$ ./e.sh 1
1
$ seq 1
1
$ ./e.sh 1 1 3
1
2
3
$ seq 1 1 3
1
2
3
$ ./e.sh 1 -1 3
$ seq 1 -1 3
$ ./e.sh 3 -1 1
3
2
1
$ seq 3 -1 1
3
2
1
$ ./e.sh 1 0 1
Step is zero!
$ seq 1 0 1|head
1
1
1
... (endlessly)
$ ./e.sh 1 .1 1.2
./e.sh: line 3: 1.2+0: syntax error: invalid arithmetic operator (error token is ".2+0")
$ seq 1 .1 1.2
1.0
1.1
1.2
Bash arithmetic is not working with floating point numbers, but seq does. If You need this feature You could use the x=$(echo $x + $step|bc) format. Also $((...)) should replaced by $(...|bc) like lines.

Answer 4

In Bash 4 you can also do it like this:
for i in {$start..$stop..$step}; do
echo $i
done