Re: Isn't this tail recursive?

Oops, I made a false statement:

...

f $! a = f a

but the difference is that $! causes "a" to be reduced completely, so it won't build a huge thunk.

This isn't true. $! will only perform one reduction, so for instance:

id $! (a+1,b+1)

will not cause a+1 and b+1 to be calculated; it will only perform the computation which creates the tuple. similarly,

id $! [a+5]

will not cause a+5 to be calculated, it will only result in the list being created (i.e., reduced from a computation which will compute [a+5] to simply the value [a+5]). if you want what i was talking about, use the DeepSeq module (http://www.isi.edu/~hdaume/haskell/Util/DeepSeq.hs) and then you can write:

id $!! [a+5]

which will actually perform the calculation. - Hal