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copied!<p>I'm unable to test, but I guess the solution is to move the <code>repo_child</code> stream into <code>slurp_repos</code>, that is:</p> <pre><code>fn main() { let (db_child, repo_child) = DuplexStream(); do spawn { slurp_repos(repo_child); } } fn slurp_repos(chan: DuplexStream<~str, ~str>) { ... do request.begin |event| { ... chan.send(api_url); } } </code></pre> <p>By moving the whole endpoint, it allows it to be transferred across tasks (because a <code>DuplexStream</code> is <code>Send</code>able). Also, note that sharing (what the use of a references allows) the endpoint of a basic bidirectional stream (which is what <code>DuplexStream</code> is) doesn't really make sense: there can only be one person at each end of a telephone.</p> <p>The error message about <code>repo_child</code> not living long enough is because the type of <code>slurp_repos</code> requires something that is <code>'static</code> i.e. lasts for the whole lifetime of the program, but <code>repo_child</code> definitely doesn't: it is local to a function.</p> <p>The reason the compiler tells you to put <code>'static</code> on <code>slurp_repos</code>, is because the only <code>Send</code>able reference is one with this lifetime. This restriction is required because you could have the owning task finish before the borrowing task does, and then destroy/deallocate the reference, leaving a dangling pointer; in diagrams:</p> <pre><code>start program/call main | v allocate repo_child | v spawn -----------------> slurp_repos(chan = &repo_child) | | v v finish do things with the chan reference to repo_child | | v v deallocate repo_child general work ... | v do more things with chan: oops, already freed; use-after-free bug </code></pre>

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