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Arithmetic evaluation of negative numbers with base prefix

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  Arithmetic evaluation of negative numbers with base prefix Jeremy Townshend <jeremy.townshend@gmail.com> - 2019-06-14 15:19 +0100

#15048 — Arithmetic evaluation of negative numbers with base prefix

Configuration Information [Automatically generated, do not change]:
Machine: x86_64
OS: linux-gnu
Compiler: gcc
Compilation CFLAGS:  -DPROGRAM='bash' -DCONF_HOSTTYPE='x86_64' -DCONF_OSTYPE='linux-gnu' -DCONF_MACHTYPE='x86_64-pc-linux-gnu' -DCONF_VENDOR='pc' -DLOCALEDIR='/usr/share/locale' -DPACKAGE='bash' -DSHELL -DHAVE_CONFIG_H   -I.  -I../. -I.././include -I.././lib  -Wdate-time -D_FORTIFY_SOURCE=2 -g -O2 -fdebug-prefix-map=/build/bash-vEMnMR/bash-4.4.18=. -fstack-protector-strong -Wformat -Werror=format-security -Wall -Wno-parentheses -Wno-format-security
uname output: Linux tower 4.15.0-48-generic #51-Ubuntu SMP Wed Apr 3 08:28:49 UTC 2019 x86_64 x86_64 x86_64 GNU/Linux
Machine Type: x86_64-pc-linux-gnu

Bash Version: 4.4
Patch Level: 19
Release Status: release

Description:
	Unexpected and undocumented behaviour for arithmetic evaluation of negative numbers when prefixed with the optional "base#" (e.g. 10#${i}). The base prefix may be needed if the variable has a decimal integer value but might be zero-padded, otherwise it is at risk of being misinterpreted as an octal.  Where the variable holds a negative value, results are as you would expect (e.g. i=-1; echo $((10#${i})), returns -1) until you subtract (or unary minus) the variable.  This unexpected behaviour occurs even when numbers are used directly (as in the first part of the Repeat-By section to simplify) but in real world examples the number would be hidden in a variable requiring the optional "base#" prefix to ensure correct interpretation of its value. 

Repeat-By:
	echo $((10#-1))   # -1 as expected
	echo $((0-10#1))  # -1 as expected
	echo $((0+10#-1)) # -1 as expected
	echo $((0-10#-1)) # -1 UNEXPECTED. Would expect 1.
	echo $((0--1))    # 1 as expected

	# Real world example:
	i=001
	echo $((3-10#${i})) # 2 as expected
	i=$((10#${i}-2))		# i's value decremented by 2 to -1
	echo $((3-10#${i})) # 2 UNEXPECTED. Would expect 4.
	echo $((3+10#${i})) # 2 as expected
	# Certainly wouldn't expect the last two expressions to have the same
	# result.

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