assert_exec cannot assert a non-zero exit code under --strict #1207

Problem

function test_x() { assert_exec "./fails.sh" --exit 1; }   # ./fails.sh exits 1

A successful command is fine in both modes, which is why this went unnoticed:
only the failing case breaks — and checking a failing command is the whole
point of --exit 1.

Cause

eval "$cmd" >"$stdout_file" 2>"$stderr_file"
local exit_code=$?

--strict enables set -e, so a non-zero eval aborts the test function
right there and local exit_code=$? never runs. Both the stdin and non-stdin
branches have it.

Found via the docs

docs/common-patterns.md → "Testing Failure Cases" shows two forms, and under
--strict both failed:

assert_exec "./src/validate_email.sh invalid-email" --exit 1   # ✗ Error
./src/validate_email.sh invalid-email; assert_general_error     # ✗ Error

The second is the $?-capture trap already documented in #1170. The first was
this bug. With it fixed, the recommended form works under --strict, so the
guide's advice stands as written.

Fix

local exit_code=0 then eval … || exit_code=$?, in both branches. Declaring
and assigning together would mask the status behind local's own.