Converting string to integer in BASH

gnashley · 04-24-2010, 11:59 AM

The portable way to do the math is like this:
var=$(($var-1))

These are all less portable:
var=((var-1))
var=$((var-1))
((var--))
let...

EricAiken · 08-31-2010, 07:23 PM

Ok, can someone explain this one?

$ a="06"
$ b="1"
$ c=$((a+b))
$ echo $c
7

$ a="07"
$ b="1"
$ c=$((a+b))
$ echo $c
8

$ a="08"
$ b="1"
$ c=$((a+b))
-sh: 08: value too great for base (error token is "08")

$ a="09"
$ b="1"
$ c=$((a+b))
-sh: 09: value too great for base (error token is "09")

$ a="10"
$ b="1"
$ c=$((a+b))
$ echo $c
11

jay73 · 08-31-2010, 08:38 PM

Quote:

$ a="08"
$ b="1"
$ c=$((a+b))
-sh: 08: value too great for base (error token is "08")

$ a="09"
$ b="1"
$ c=$((a+b))
-sh: 09: value too great for base (error token is "09")

Prefacing a number with 0 makes it octal. Your problem is that octal uses units of 8, expressed as the range 0-7, so 08 is plain nonsense. Decimal 8 would be octal (0)10 and decimal 9 would be octal (0)11.

echo $((8#10)) <--- base 8 (octal) value 10 is decimal 8; this will print 8
echo $((2#10)) <--- base 2 (binary) 10 is decimal 2; it will print 2
echo $((16#10)) <--- base 16 (hexadecimal) 10 is decimal 16; it will print 16