Configuration Information [Automatically generated, do not change]:
Machine: i586
OS: linux-gnu
Compiler: gcc -I/usr/src/packages/BUILD/bash-3.2
-L/usr/src/packages/BUILD/bash-3.2/../readline-5.2
Compilation CFLAGS: -DPROGRAM='bash' -DCONF_HOSTTYPE='i586'
-DCONF_OSTYPE='linux-gnu' -DCONF_MACHTYPE='i586-suse-linux-gnu'
-DCONF_VENDOR='suse' -DLOCALEDIR='/usr/share/locale' -DPACKAGE='bash' -DSHELL
-DHAVE_CONFIG_H -I. -I. -I./include -I./lib -march=i586 -mtune=i686
-fmessage-length=0 -O2 -Wall -D_FORTIFY_SOURCE=2 -fstack-protector
-funwind-tables -fasynchronous-unwind-tables -g -D_GNU_SOURCE -DRECYCLES_PIDS
-Wall -g -std=gnu89 -Wextra -Wno-unprototyped-calls -Wno-switch-enum -pipe
-fPIE -fprofile-use
uname output: Linux spatzenhorst 2.6.27.23-0.1-pae #1 SMP 2009-05-26 17:02:05
-0400 i686 i686 i386 GNU/Linux
Machine Type: i586-suse-linux-gnu
Bash Version: 3.2
Patch Level: 39
Release Status: release
Description:
The implementation of "set -e" does not respect "lexical nesting".
This can be very surprising. The documentation should explicitly
and clearly explain the behavior.
In a context of "set -e", with function "f() { x; false; y; }",
"f" will run only "x", while "f || true" will also run "y".
This is highly surprising.
The documentation states that "shell does not exit if the command that
fails is part of [a boolean expression]" (paraphrased). The problem
here is the phrase "is part of". At first glance, this appears to mean
"is lexically part of", because the wording refers to a number of
syntactical constructs. This is the relevant sentence in full:
The shell does not exit if the command that fails is part of the
command list immediately following a while or until keyword, part of
the test in an if statement, part of a && or || list, or if the
command's return value is being inverted via !.
However, "is part of" apparently does not mean "is lexically part of".
Instead, it means "is executed in the execution context of", as in "is
executed when a preceding stack frame represents".
Repeat-By:
set -e; f() { echo before; false; echo after; }; f || true; echo rc=$?;
prints:
----
before
after
rc=0
----
while many if not most programmers would expect it to print
----
before
rc=0
----
Fix:
If the effect is intended, clearly document it.
If not, change the implementation.
Note that some other shells, in particular ksh, implement the same
behavior.
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