Addition in bash

[David N. Lombard]

Ray Parrish wrote:
> Hello,
> 
> I have the following script fragment
> 
> Seconds=0;
> while [ "$Seconds" -lt "60" ]
>   do
>   sleep 5;
>   Seconds=$Seconds+5;
>   padsp espeak "$Seconds"
> done
> 
> For some reason Seconds becomes "0+5" instead of the integer value 5 
> which is what I am attempting to get it to be.

Yes, you're doing a string concatenation there, "+5" is being 
concatenated to "0" to get "0+5". Exactly what you asked for. ;)

> Does anyone know how to do integer math in bash?

For *bash* and not sh, I would rewrite the above as:

   Seconds=5
   while [[ $Seconds < 60 ]]
   do
     sleep 5
     (( Seconds += 5 ))
     padsp espeak $Seconds
   done

Note these changes:

1) I used the *bash* double bracket conditional form.  This offers 
better testing *and* it changes how the strings are interpreted, for 
example, you don't have to quote null strings or protect special 
characters since you're not building an executable string.

2) The double parentheses arithmetic expression.  This doesn't require 
the $ to indicate a variable and permits C-style operators.

3) You don't have to quote every occurrence of every string.  You only 
need to be mindful of spaces and special characters.  So, I didn't quote 
the $Seconds on the padsp line.

4) You only need the semicolon to break a single physical line into 
multiple statements.  You don't have that here.  An example that would 
require the semicolon is

   while [[ $Seconds < 60 ]] ; do

BTW, you may have meant to initialize $Seconds at 0.  The current form 
will result in 10, 15, 20, 25, ... 60.

Also, you could do this without arithmetic using

   for Seconds in $(seq 5 5 60) ; do
     sleep $Seconds
     padsp espeak $Seconds
   done

Note that seq puts the increment before the last value, unlike most for 
constructs.

If you really did want sh, Bourne shell, you would do the arithmetic as

   Seconds=`expr $Seconds + 5`

-- 
David N. Lombard