I'm puzzled by bash (and dash) behavior when
-e option is set.
false && true
false && true
While first occurrence works as expected, second occurrence (inside function) leads to immediate exit. I searched documentation but was unable to find explanation to such different behavior. Is there any rationale behind this, or is this bug? I tested this in bash and dash with same results.
According to bash manpage:
-eExit immediately if a pipeline (which may consist of a single simple command), a list, or a compound command (see SHELL GRAMMAR above), exits with a non-zero status. The shell does not exit if the command that fails is part of the command list immediately following a
untilkeyword, part of the test following the
elifreserved words, part of any command executed in a
||list except the command following the final
||, any command in a pipeline but the last, or if the command's return value is being inverted with
!. If a compound command other than a subshell returns a non-zero status because a command failed while
-ewas being ignored, the shell does not exit. A trap on
ERR, if set, is executed before the shell exits. This option applies to the shell environment and each subshell environment separately (see COMMAND EXECUTION ENVIRONMENT above), and may cause subshells to exit before executing all the commands in the subshell.
If a compound command or shell function executes in a context where
-eis being ignored, none of the commands executed within the compound command or function body will be affected by the
-esetting, even if
-eis set and a command returns a failure status. If a compound command or shell function sets
-ewhile executing in a context where
-eis ignored, that setting will not have any effect until the compound command or the command containing the function call completes.