bash: $? after a $(...) in the same word is the substitution's status

In cmd; echo "$(date) rc=$?", bash expands left to right, so $? is read after $(date) has already run and gives date's exit status (0), not cmd's. Tested on bash 5.2.21: false; echo "$(date +%s) rc=$?" printed rc=0, while false; echo "rc=$? $(date +%s)" printed rc=1. Save it first (cmd; r=$?; echo "$(date) rc=$r"), which printed rc=1. Logging wrappers that timestamp a line with $(date) are where this usually bites.