Hi Wolfgang, you are right, this behaviour is a bit surprising. What you want to observe can be obtained with, for example: loop c = putChar c >> loop c where loop is not a CAF anymore since it takes an argument. In the definition that you gave, main is a CAF and gets evaluated only the first time. The subsequent times it is entered do not trigger any further evaluation: the IO action is already evaluated and the only thing left to do is execute it. This does not trigger a breakpoint since no evaluation proper is done. For more information about CAFs see http://haskell.org/haskellwiki/Constant_applicative_form That page remarks that CAFs are an optimization which does not alter referential transparency. Unfortunately their special behaviour is highlighted by the ghci debugger. If you want to find the reason for an infinite loop: 1. enable the flag -fbreak-on-error 2. run your expression with :trace (eg :trace loop 'a') 3. hit Ctrl-C while your program is stuck in the loop to have the debugger break in the loop. 4: use :history and :back to find out where the loop is located and why. Cheers pepe On 20/11/2007, at 16:09, Wolfgang Jeltsch wrote:
Hello,
please consider the following program:
main = putChar 'A' >> main
I load this into GHCi, enter :step main, followed by :step and a second :step. Although I would expect to get the As in the output step by step, GHCi now hangs inside the infinite loop. Is this intentional? To me, this seems rather weird. I wanted to use GHCi’s debugger to find the reason for an infinite recursion but from my experience it looks as if this might not be possible. :-(
Best wishes, Wolfgang _______________________________________________ Glasgow-haskell-users mailing list Glasgow-haskell-users@haskell.org http://www.haskell.org/mailman/listinfo/glasgow-haskell-users