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    <p>This doesn't stack overflow - even in the interpreter, where there are no optimizations and no rewrite rules - because it doesn't use the stack.</p> <p>Look at the definition of (++), for example,:</p> <pre><code>(++) :: [a] -&gt; [a] -&gt; [a] (++) [] ys = ys (++) (x:xs) ys = x : xs ++ ys </code></pre> <p>The key thing is <code>x : (xs ++ ys)</code> -- that is, it is recursion guarded by the (:) "cons" constructor. Because Haskell is lazy, it allocates a thunk for the cons operation, and the recursive call goes onto this (heap-allocated) thunk. So your stack allocation is now heap allocation, which can expand quite a bit. So all this does is walk the big list, allocating new cons objects on the heap to replace the ones it is traversing. Easy!</p> <p>"reverse" is a bit different:</p> <pre><code>reverse l = rev l [] where rev [] a = a rev (x:xs) a = rev xs (x:a) </code></pre> <p>That is a tail recursive, accumulator-style function, so again, it will allocate on the heap.</p> <p>So you see, the functions rely on using cons cells on the heap, instead of on the stack, hence no stack overflow.</p> <p>To really nail this, look at the runtime stats from the GC vm:</p> <pre><code>$ time ./B +RTS -s 99 833 MB total memory in use (13 MB lost due to fragmentation) Generation 0: 3054 collections, 0 parallel, 0.99s, 1.00s elapsed %GC time 82.2% (85.8% elapsed) </code></pre> <p>There's your big list -- it is allocated on the heap, and we spend 80% of the time cleaning up cons nodes that are created by (++).</p> <p>Lesson: you can often trade stack for heap.</p>
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    1. USNorman Ramsey
    1. USDon Stewart
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    1. COGreat answer, thanks! Love your book, btw, great work!
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    2. CO+1 for being the only person on SO who actually understands lazy evaluation. My early training in ML has left me permanently crippled :-(
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