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    <p>There are two ways to do this without using concurrency, both with caveats.</p> <p>The first way is that if <code>pipe1</code> and <code>pipe2</code> are just simple <code>Consumer</code>s that loop forever like:</p> <pre><code>p1 = for cat f -- i.e. p1 = forever $ await &gt;&gt;= f p2 = for cat g -- i.e. p2 = forever $ await &gt;&gt;= g </code></pre> <p>... then the easy way to solve this is to just write:</p> <pre><code>for P.stdinLn $ \str -&gt; do f str g str </code></pre> <p>For example, if <code>p1</code> is just <code>print</code>ing every value:</p> <pre><code>p1 = for cat (lift . print) </code></pre> <p>... and <code>p2</code> is just writing that value to a handle:</p> <pre><code>p2 = for cat (lift . hPutStrLn h) </code></pre> <p>... then you would combine them like so:</p> <pre><code>for P.stdinLn $ \str -&gt; do lift $ print str lift $ hPutStrLn h str </code></pre> <p>However, this simplification only works for <code>Consumer</code>s that trivially loop. There's another solution that is more general, which is to define an <code>ArrowChoice</code> instance for pipes. I believe that pull-based <code>Pipe</code>s do not permit a correct law-abiding instance, but push-based <code>Pipe</code>s do:</p> <pre><code>newtype Edge m r a b = Edge { unEdge :: a -&gt; Pipe a b m r } instance (Monad m) =&gt; Category (Edge m r) where id = Edge push (Edge p2) . (Edge p1) = Edge (p1 &gt;~&gt; p2) instance (Monad m) =&gt; Arrow (Edge m r) where arr f = Edge (push /&gt;/ respond . f) first (Edge p) = Edge $ \(b, d) -&gt; evalStateP d $ (up \&gt;\ unsafeHoist lift . p /&gt;/ dn) b where up () = do (b, d) &lt;- request () lift $ put d return b dn c = do d &lt;- lift get respond (c, d) instance (Monad m) =&gt; ArrowChoice (Edge m r) where left (Edge k) = Edge (bef &gt;=&gt; (up \&gt;\ (k /&gt;/ dn))) where bef x = case x of Left b -&gt; return b Right d -&gt; do _ &lt;- respond (Right d) x2 &lt;- request () bef x2 up () = do x &lt;- request () bef x dn c = respond (Left c) </code></pre> <p>This requires a newtype so that the type parameters are in the order that <code>ArrowChoice</code> expects.</p> <p>If you're unfamiliar with the term push-based <code>Pipe</code>, it's basically a <code>Pipe</code> that begins from the most upstream pipe instead of the most downstream pipe, and they all have the following shape:</p> <pre><code>a -&gt; Pipe a b m r </code></pre> <p>Think of it as a <code>Pipe</code> that cannot "go" until it receives at least one value from upstream.</p> <p>These push-based <code>Pipe</code>s are the "dual" to conventional pull-based <code>Pipe</code>s, complete with their own composition operator and identity:</p> <pre><code>(&gt;~&gt;) :: (Monad m) =&gt; (a -&gt; Pipe a b m r) -&gt; (b -&gt; Pipe b c m r) -&gt; (a -&gt; Pipe a c m r) push :: (Monad m) -&gt; a -&gt; Pipe a a m r </code></pre> <p>... but the unidirectional <code>Pipes</code> API does not export this by default. You can only get these operators from <code>Pipes.Core</code> (and you may want to study that module more closely to build an intuition for how they work). That module shows that both push-based <code>Pipe</code>s and pull-based <code>Pipe</code>s are both special cases of more general bidirectional versions, and understanding the bidirectional case is how you learn why they are duals of each other.</p> <p>Once you have an <code>Arrow</code> instance for push-based pipes, you can write something like:</p> <pre><code>p &gt;&gt;&gt; bifurcate &gt;&gt;&gt; (p1 +++ p2) where bifurcate = Edge $ pull ~&gt; \a -&gt; do yield (Left a) -- First give `p1` the value yield (Right a) -- Then give `p2` the value </code></pre> <p>Then you would use <code>runEdge</code> to convert that to a pull-based pipe when you are done.</p> <p>This approach has one major draw-back, which is that you can't automatically upgrade a pull-based pipe to a push-based pipe (but usually it's not hard to figure out how to do it manually). For example, to upgrade <code>Pipes.Prelude.map</code> to be a push-based <code>Pipe</code>, you would write:</p> <pre><code>mapPush :: (Monad m) =&gt; (a -&gt; b) -&gt; (a -&gt; Pipe a b m r) mapPush f a = do yield (f a) Pipes.Prelude.map f </code></pre> <p>Then that has the right type to be wrapped up in the <code>Arrow</code>:</p> <pre><code>mapEdge :: (Monad m) =&gt; (a -&gt; b) -&gt; Edge m r a b mapEdge f = Edge (mapPush f) </code></pre> <p>Of course, an even simpler way would be just to write it from scratch:</p> <pre><code>mapEdge f = Edge $ push ~&gt; yield . f </code></pre> <p>Use whichever approach suits you best.</p> <p>In fact, I came up with the <code>Arrow</code> and <code>ArrowChoice</code> instances precisely because I was trying to answer the exact same question as you: how do you solve these kinds of problems without using concurrency? I wrote up a long answer about this more general subject in another Stack Overflow answer <a href="https://stackoverflow.com/questions/19201901/in-functional-reactive-programming-how-do-you-share-state-between-two-parts-of/19204253#19204253">here</a>, where I describe how you can use these <code>Arrow</code> and <code>ArrowChoice</code> instances to distill concurrent systems into equivalent pure ones.</p>
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