Some people write assembly language by hand. Some of them have a good reason to do so. But there's no good reason to write Nock by hand, ever.
Except for the purposes of learning Nock. It's not necessary to know Nock to program in Hoon. But it's quite desirable. The feeling of understanding the system at two levels is pleasant.
Decrement in Nock
A good practice exercise for Nock is a decrement formula. Ie, a formula f which implements the partial function that produces (s - 1) if s is a nonzero atom, and otherwise does not terminate.
As we know, the equivalent formula for increment is
[4 0 1]
Thus:
~>zod:dojo .*(42 [4 0 1])43
Of course, increment is built into Nock. So, ha, that's easy.
The best way to learn Nock is to stop reading right now, and go write your own decrement formula. There's no substitute for doing it yourself. But the second best way is faster...
Decrement in Hoon
How do we decrement? A good way to start is to gaze fondly on how we'd do it if we actually had a real language, ie, Hoon. Here is a minimal decrement in Hoon:
=> a=. :: line 1=+ b=0 :: line 2|- :: line 3?: =(a +(b)) :: line 4b :: line 5$(b +(b)) :: line 6
Or for fun, on one line:
=>(a=. =+(b=0 |-(?:(=(a +(b)) b $(b +(b))))))
Does Hoon actually work?
~zod:dojo> =>(42 =>(a=. =+(b=0 |-(?:(=(a +(b)) b $(b +(b)))))))41
Let's translate this into English. How do we decrement the subject? First (line 1), we rename the subject a. Second (line 2), we add a variable, b, an atom with value 0. Third (line 3), we loop. Fourth, we test if a equals b plus 1 (line 4), produce b if it does (line 5), repeat the loop with b set to b plus 1 (line 6) if it doesn't. Obviously, while the syntax is unusual, the algorithm is anything but deep. We are calculating b minus one by counting up from 0.
(Obviously, this is an O(n) algorithm. Is there a better way? There is not. Do we actually do this in practice? Yes and no.)
Unfortunately we are missing a third of our Rosetta stone. We have decrement in Hoon and we have it in English. How do we express this in Nock? What will the Hoon compiler generate from the code above? Let's work through it line by line.
Nock has no types, variable names, etc. So line 1 is a no-op.
How do we add a variable (line 2)? We compute a new subject, which is a cell of the present subject and the variable. With this new subject, we execute another formula.
Since 0 is a constant, a formula that produces it is
[1 0]
To combine 0 with the subject, we compute
[[1 0] [0 1]]
which, if our subject is 42, gives us
[0 42]
which we can use as the subject for an inner formula, g. Composing our new variable with g, we have f as
[2 [[1 0] [0 1]] [1 g]]
which seems a little funky for something so simple. But we can simplify it with the composition macro, 7:
[7 [[1 0] [0 1]] g]
and still further with the augmentation macro, 8:
[8 [1 0] g]
If you refer back to the Nock definition, you'll see that all these formulas are semantically equivalent.
Let's continue with our decrement. So what's g? We seem to loop. Does Nock have a loop instruction? It most certainly does not. So what do we do?
We build a noun called a core - a construct which is behind any kind of interesting control flow in Hoon. Of course, the Nock programmer is not constrained to use the same techniques as the Hoon compiler, but it is probably a good idea.
In Hoon, all the flow structures from your old life as an Earth programmer become cores. Functions and/or closures are cores, objects are cores, modules are cores, even loops are cores.
The core is just a cell whose tail is data (possibly containing other cores) and whose head is code (containing one or more formulas). The tail is the payload and the head is the battery. Hence your core is
[bat pay]
To activate a core, pick a formula out of the battery, and use the entire core (not just the payload) as the subject.
(A core formula is called an arm. An arm is almost like an object-oriented method, but not quite - a method would be an arm that produces a function on an argument. The arm is just a function of the core, ie, a computed attribute.)
Of course, because we feed it the entire core, our arm can invoke itself (or any other formula in the battery). Hence, it can loop. And this is what a loop is - the simplest of cores.
We need to do two things with this core: create it, and activate it. To be precise, we need two formulas: a formula which produces the core, and one which activates its subject. We can compose these functions with the handy 7 instruction:
[8 [1 0] [7 p a]]
p produces our core, a activates it. Let's take these in reverse order. How do we activate a core?
Since we have only one formula, it's the battery itself. Thus we want to execute Nock with the whole core (already the subject, and the entire battery (slot 2)). Hence, a is
[2 [0 1] [0 2]]
We could also use the handy 9 macro - which almost seems designed for firing arms on cores:
[9 2 [0 1]]
Which leaves us seeking
[8 [1 0] [7 p [9 2 0 1]]]
And all we have to do is build the core, p. How do we build a core? We add code to the subject, just as we added a variable above. The initial value of our counter was a constant, 0. The initial (and permanent) value of our battery is a constant, the loop formula l. So p is
[8 [1 l] [0 1]]
Which would leave us seeking
[8 [1 0] [7 [8 [1 l] [0 1]] [9 2 0 1]]]
except that we have duplicated the 8 pattern again, since we know
[7 [8 [1 l] [0 1]] [9 2 0 1]]
is equivalent to
[8 [1 l] [9 2 0 1]]
so the full value of f is
[8 [1 0] [8 [1 l] [9 2 0 1]]]
Thus our only formula to compose is the loop body, l. Its subject is the loop core:
[bat pay]
where bat is just the loop formula, and pay is the pair [b a], b being the counter, and a the input subject. Thus we could also write this subject as
[l b a]
and we see readily that a is at slot 7, b 6, l 2. With this subject, we need to express the Hoon loop body
?: =(a +(b)) :: line 4b :: line 5$(b +(b)) :: line 6
This is obviously an if statement, and it calls for 6. Ie:
[6 t y n]
Giving our decrement program as:
[8 [1 0] [8 [1 6 t y n] [9 2 0 1]]]
For t, how do we compute a flag that is yes (0) if a equals b plus one? Equals, we recall, is 5. So t can only be
[5 [0 7] [4 0 6]]
If so, our product y is just the counter b:
[0 6]
And if not? We have to re-execute the loop with the counter incremented. If we were executing it with the same counter, obviously an infinite loop, we could use the same core:
[9 2 0 1]
But instead we need to construct a new core with the counter incremented:
[l +(b) a]
ie,
[[0 2] [4 0 6] [0 7]]
and n is:
[9 2 [[0 2] [4 0 6] [0 7]]]
Hence our complete decrement. Let's reformat vertically so we can actually read it:
[8[1 0][ 8[ 1[ 6tyn]][9 2 0 1]]]
which becomes
[8[1 0][ 8[ 1[ 6[5 [0 7] [4 0 6]][0 6][9 2 [[0 2] [4 0 6] [0 7]]]]][9 2 0 1]]]
or, on one line without superfluous brackets:
[8 [1 0] 8 [1 6 [5 [0 7] 4 0 6] [0 6] 9 2 [0 2] [4 0 6] 0 7] 9 2 0 1]
which works for the important special case, 42:
~zod:dojo> .*(42 [8 [1 0] 8 [1 6 [5 [0 7] 4 0 6] [0 6] 9 2 [0 2] [4 0 6] 0 7] 9 2 0 1])41
If you understood this, you understand Nock. At least in principle! Now, go write an adder. Or don't.