bash - How does "set -e" work with subshells?

Question

I was wondering whether set -e propagates through subshells (i.e. does a subshell inherit the -e setting of its parent), so I made some experiments. I found some strange results that I can't explain.

First, here are some basic tests. They return what I expect.

( true; false )         # 1
( false; true )         # 0
( set -e; false; true ) # 1

Now I tried what happens if I put a subshell within my subshell. This expression returns 1, which suggests that it propagates.

( set -e; ( false; true ) )

Then I tried these expressions. I expected them to return 1, but I found that they return 0.

( set -e; ( true; false ); true )
( set -e; ( set -e; false; true ); true )

Why? In both cases, the inner subshell returns 1, whether set -e propagates or not (as I checked in the beginning). The outer subshell has set -e, which means that it should fail after the inner subshell exits, but it does not. Can someone explain this?

Answer 1

Prior to bash 4, set -e appears to only cause the shell to exit if a simple command has a non-zero exit (emphasis mine):

-e Exit immediately if a simple command (see SHELL GRAMMAR above) exits with a non-zero status.

In bash 4 (possibly 4.1, I don't have a 4.0 to check), the effect of -e was extended to more complicated commands:

-e Exit immediately if a pipeline (which may consist of a single simple command), a subshell com- mand enclosed in parentheses, or one of the commands executed as part of a command list enclosed by braces (see SHELL GRAMMAR above) exits with a non-zero status.