On Wed, Jan 23, 2008 at 11:28:35PM -0800, Brad Beyenhof wrote:
Yes, normal order attempts to evaluate the operator and all operands to primitive values before performing any sort of substitution. Since (p) always evaluates back to (p) (a non-primitive), that is why it will hang.
Although you're swapping the terms normal and applicative order. Scheme is applicative order. - applicative order: evaluate arguments and apply - normal order: substitute arguments and reduce Almost all programming languages are applicative order. Dave -- [email protected] http://www.kernel-panic.org/cgi-bin/mailman/listinfo/kplug-lpsg
