Q:

Difference between normal-order and applicative-order evaluation

The language I'm learning is Scheme and I'm working on an exercise that gives this:
(define (p) (p) )

(define (test x y) (if (= x 0)
                               0
                               y))

Then the question asks to evaluate the expression: (test 0 (p) ) and to comment on the behavior that would be observed under normal - order and applicative order evaluation.
These are my thoughts:
Under normal order, the program would evaluate the subexpressions before proceeding:
Thus ( test 0 (p) ) becomes:
(test 0 p)
( if (= x 0) 0 p))

Returning output 0
The only difference in applicative order would be that the program would run as: 
( test 0 (p) ) becomes:
(test 0 (p))
( if (= x 0) 0 (p)))

Returning output 0 
Thus (p) would never be evaluated as it won't be needed.
Is this correct? Please let me know as I have almost no experience programming. 

A:

What research have you done to answer that question? I just plugged it
as it is in Google, and got as second answer (the first may be as
good, I did not check) a reference to a section of a bible on your
topic: Hal Abelson's, Jerry Sussman's and Julie Sussman's Structure
and Interpretation of Computer Programs (MIT Press, 1984; ISBN
0-262-01077-1), aka the wizard book. The reference is to the section
"Normal Order and Applicative Order".
It states:

Scheme is an applicative-order language, namely, that all the arguments to Scheme procedures are evaluated when the procedure is applied. In contrast, normal-order languages delay evaluation of procedure arguments until the actual argument values are needed.

and adds that the latter is called lazy evaluation.
So you have your definitions wrong, and taking one for the other:

applicative order evaluates subexpressions  when, i.e. just before,
the procedure is applied.
normal order passes subexpressions as they are, without evaluation,
and proceeds with the evaluation only when the corresponding formal
parameter is actually to be itself evaluated. (there is a further
twist to it regarding environment issues ... but we better forget
that at this point).

Furthermore, you do not properly understand the call mechanism which
involves two things:

the parameter passing mechanism, which include proper processing of
actual arguments to the call, depending on the evaluation rule;
the "replacement" of the call to the function by the body of the
function (without the header).

In the case of applicative order evaluation of ( test 0 (p) ), you are
supposed to evaluate the argument subxpressions first. These are 0
and (p).

evaluation of a literal value like 0 yield that value.
the second argument however is a procedure call to a parameter-less
procedure called p. It has no parameter, so that we have no worry
about evaluation order. Then, in order to pursue evaluation, we
have to replace the call by the body of the procedure which follows
the list of arguments, and then evaluate that body. The body of
procedure p, as defined by the declaration (define (p) (p) ),
is (p), so that we are left with the evaluation of what we were
just trying to evaluate. In order words, the evaluation process is
caught in a loop, and will not terminate.

... and you never get to actually finish the call to the function test,
since evaluation of its arguments does not terminate. Your program
does not terminate.
This risk of non termination, even when the guilty argument will never
be used in the call, is one of the reasons to use instead normal
order evaluation, which may be a bit harder to implement, but may have
better termination properties.
Under normal order evaluation, you do not touch the argument
sub-expressions. What you do is replace the call ( test 0 (p) ) by
the body of the function test, i.e. (if (= x 0) 0 y), where the
names of the (formal) arguments x and y are replaced by the
corresponding actual arguments 0 and (p) (up to environment, or
renaming issues, that are important but would complicate the
explanation here, and are the main difference between the original Lisp and
the Scheme language).
Hence you replace the evaluation of ( test 0 (p) ) by the evaluation
of (if (= 0 0) 0 (p)).
Now the function if is a primitive function that usually always
evaluate its first argument, but evaluates its last 2 arguments in
normal order, evaluating only the useful one, depending on whether the
first evaluates to false or true (actually NIL or #f for
false, or some other value for true, in the case of Scheme - if my
memory does not fail me). Since (= 0 0) evaluates to true,
evaluation of the conditional amount to evaluating the yet unevaluated
second argument, which is 0, which unsurprisingly (except in old
Fortran) evaluates to 0.
Deep breath.

