Matthew Brecknell wrote: > The key point of the example is that foldl itself doesn't need any of > the intermediate values of the accumulator, so these just build up into > a deeply-nested unevaluated thunk. When print finally demands an > integer, the run-time pushes a stack frame for each level of parentheses > it enters as it tries to evaluate the thunk. Too many parentheses leads > to a stack overflow. Of course, the solution to the example is to use
What is the point in building this huge thunk if it can't be evaluated without a stack overflow? Could the runtime do partial evaluation to keep the thunk size down or would that cause semantic breakage? Joe Buehler _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
