Behaviour of "eval" under "set -e" in conditional expression

That no other shell needs such workaround is an strong indication that it is a bug in OpenBSD ksh. In fact, ksh93 doesn't show such issue.

That there is a || in the command line must avoid the shell exit caused by an return code of 1 on the left side of it.

The error of an special built-in shall cause the exit of a non interactive shell acording to POSIX but that is not always true. Trying to continue out of a loop is an error, and continue is a builtin. But most shells do not exit on:

continue 3

A builtin that emits a clear error but doesn't exit.

So, the exit on false is generated by the set -e condition not by the builtin characteristic of the command (eval in this case).

The exact conditions on which set -e shall exit are quite more fuzzy in POSIX.

[sorry if this is not a real answer, I'll update it when I get round to it]

I had a look at the source code, and my conclusions are:

1) It's a bug/limitation, nothing philosophical behind it.

2) The "fix" for it from the portable fork of OpenBSD's ksh (mksh) is very poor, only making things worse, without really fixing it:

New bug, different from all the other shells:

mksh -ec 'eval "false; echo yup"'
yup

bash -ec 'eval "false; echo yup"'
(nothing)

Still not really fixed:

mksh -ec 'eval "set -e; false" || echo yup'
(nothing)

bash -ec 'eval "set -e; false" || echo yup'
yup

You can replace bash above with dash, zsh, yash, ksh93, etc.