unusual bash array command

Question

Found this snippet of code which uses an array in Bash.

unset var1[@]

for i in {0..23}
do
    var1[10#$i]="some data"
done

When setting the element with data, what's that 10# in front of the element number for?

I know that for showing the total number of elements you can use echo ${#var1[@]} but I don't think the hash in setting the element has something to do with total size.

choroba · Accepted Answer · 2019-05-07 11:16:24Z

6

In arithmetic expressions, you can use # to specify base:

$ echo $((2#1110))
14

Using 10#$i would prevent i=08 from failing with value too great for base and interpret it as 8. For 0 .. 23, it's equivalent to just i.

answered May 7, 2019 at 11:16

choroba

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4 Comments

captcha Over a year ago

That is bizarre.. I never would have guessed it. Yup, without 10# the i=07 goes fine and i=08 fails with "-bash: 08: value too great for base (error token is "08")". Whoa... So pardon my ignorance, what is the default base for defining the array element number then?

danrodlor Over a year ago

Decimal, but the shell would automatically interpret it as octal if the number starts by '0'. Is the same that the leading '0x' for hexadecimal numbers.

captcha Over a year ago

I'm gonna chmod myself outta here.. Thanks all... :-)

Wiimm Over a year ago

You can use it for bases from 2 to 64. Nice: echo $((50#A0)) will print 1800 (50*36) and echo $((64#_0)) 4032.

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