Bashscripts.org

Hello all I'm new here. I searched the web but couldn't find an answer to the following. Consider this very simple for loop.

Now if I run this script with two parameters it works perfectly. What I don't understand is when I run the script with only one parameter it still works. It will only execute the for loop once. Call me crazy but I thought for sure it would error out because $2 is unsigned. What is going on here? Is bash smart enough to drop $2 from the list because it doesn't exist? Or is this bad coding practice? I wan't my script to work on many different computers so I don't want any anomalous behavior.

Sorry if this is a silly question or if I am missing something obvious. I am new to bash and linux in general. Thx for the replies in advance!

Hi crazy,

add an -x to your shebang, and see how your loop works

bash doesn't drop $2, $2 is just considered by its value: '' (i.e. none)

if you want to treat all the parameters passed to your script, you could even omit the in $list part.


you could also do:

good, bad coding practice? the real questions are:
- does it do the job?
- if it does, how could you optimize it?

bash interprets the line by putting in the variable values first so;



where $1=one and $2 is blank the line looks like;



this also demonstrates why quoting variables is important, consider where $1="one two" and $2="three four" the line would be interpreted as;



which would actually loop four times. Using quotes;



where $1 and $2 are as in the previous example, gets interpreted to



and would only be looped twice, with param="one two" and then "three four"