At a high level, it means it does what I think you're looking for: the
"read end" of the channel created there isn't connected to anything, so
there's no leak if there's no consumers. But duplicating it gives you a new
"read end" that yields values as you would expect.
Without going into too much detail about STM, everything runs in a
transaction, and `retry` means to roll back any changes within the
transaction and start again. An unconditional `retry` (including
unconditionally reading from a channel created with `newBroadcastTChan`) is
a deadlock and therefore probably a mistake. So don't do that!
On 25 Jan 2017 05:41, "Saurabh Nanda"
box:
https://hackage.haskell.org/package/stm-2.4.4.1/docs/Control -Concurrent-STM-TChan.html#v:newBroadcastTChan
What does the following comment in the documentation really mean (highlighted by >>><<<)? "Create a write-only TChan https://hackage.haskell.org/package/stm-2.4.4.1/docs/Control-Concurrent-STM-....
More precisely, readTChan https://hackage.haskell.org/package/stm-2.4.4.1/docs/Control-Concurrent-STM-... will retry https://hackage.haskell.org/package/stm-2.4.4.1/docs/Control-Monad-STM.html#... even after items have been written to the channel.<<< The only way to read a broadcast channel is to duplicate it with dupTChan https://hackage.haskell.org/package/stm-2.4.4.1/docs/Control-Concurrent-STM-... ."
-- Saurabh.