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Old 04-24-2010, 11:59 AM   #16
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The portable way to do the math is like this:

These are all less portable:
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Old 08-31-2010, 07:23 PM   #17
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Ok, can someone explain this one?

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

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

$ 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 $c
Old 08-31-2010, 08:38 PM   #18
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$ 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

Last edited by jay73; 08-31-2010 at 08:46 PM.
2 members found this post helpful.


bash, conversion, convert, integer, string, type

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