Re: bc within perl

This is not an error in perl. Re-writing your first command directly in the shell we see this:

$ echo "ibase=16;obase=16;(A5A5A5A5A5A5A5A5/8)" | bc
 05 06 01 09 10 06 00 00 05 16 14 07 14 20
$
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If this isn't the output you expect then you will need to dig deeper into usage of bc before wrapping it in anything else. I don't use it so can't comment further.

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Re^2: bc within perl by amudelkaa (Novice) on Nov 08, 2016 at 09:40 UTC
Aah thanks. My bad. I changed it a bit. For some reason, obase variable should be defined first and then ibase. This is now fixed. Thank you for the help, monks :).	[reply]
Re^3: bc within perl by pryrt (Abbot) on Nov 08, 2016 at 14:55 UTC
For some reason, obase variable should be defined first and then ibase specifically, because you told it to output in base twenty-two. `ibase=16; # set the input base to hexadecimal. # all numerical inputs will be treated with an # implied 0x prefix obase=16; # 16 is implied hexadecimal 0x16 = sixteen plus six = twe +nty-two (A5A5A5A5A5A5A5A5/8); # divide 0xA5... by 0x8, and output in base t +wenty-two` [download] And my copy of the `bc` manpage explains "For bases 2 through 16, the usual method of writing numbers is used. For bases greater than 16, bc uses a multi-character digit method of printing the numbers where each higher base digit is printed as a base 10 number. The multi-character digits are separated by spaces. ..." Sanity check: you expected 0x14b4b4b4b4b4b4b4, which is about (0x14) * 16*14, plus some change: 20 16*14 = 1.44e18, plus. You got "05 06 01 09 10 06 00 00 05 16 14 07 14 20", which is about 05 22**13, plus some change: 1.41e18, plus. Yep, the output was in base-22.	[reply] [d/l] [select]