@@ -12,10 +12,9 @@ In this lesson, we will write a generator that takes an integer and checks if it
12|= [* [n=@ud ~] ~]
12|= [* [n=@ud ~] ~]
17− (gte n 1)
15+?& (gte n 1)
18 (lte n 100)
16 (lte n 100)
@@ -41,53 +40,32 @@ This line casts the output as a `flag`, which is a type whose values are `%.y` a
41Let's look at the conditional.
40Let's look at the conditional.
47`?&` (pronounced "wut-pam") takes in a list of Hoon expressions terminated by `==` that evaluate to a `flag` and returns the logical "AND" of these `flag`s. Most runes take a fixed number of children, but the handful that do not (such as `?&`) end their list of children with a terminating rune. In our context, that means that if the product of each of the children of `?&` is `%.y`, then the product of the entire `?&` expression is `%.y` as well. Otherwise, the product of the conditional `?&` is `%.n`.
46`?&` (pronounced "wut-pam") takes in a list of Hoon expressions terminated by `==` that evaluate to a `flag` and returns the logical "AND" of these `flag`s. Most runes take a fixed number of children, but the handful that do not (such as `?&`) end their list of children with a terminating rune. In our context, that means that if the product of each of the children of `?&` is `%.y`, then the product of the entire `?&` expression is `%.y` as well. Otherwise, the product of the conditional `?&` is `%.n`.
49−The first child of `?&` is the following.
48+The first child of `?&` is `(gte n 1)`. It is good practice to put the first boolean test of a conditional on the same line as the conditional as we have done here. This utilizes the standard library function `gte` which stands for "greater than or equal to". `(gte a b)` returns `%.y` if `a` is greater than or equal to `b`, and `%.n` otherwise.
55−This checks to see if `0` is equal to `(mod n 2)` and returns a `flag`. In other words, it produces `%.y` if `n` is even and `%.n` if `n` is odd.
57Next we have:
50Next we have:
61−This utilizes the standard library function `gte` which stands for "greater than or equal to". `(gte a b)` returns `%.y` if `a` is greater than or equal to `b`, and `%.n` otherwise.
64−(lte n 100)
52+ (lte n 100)
66`lte` is the standard library function for "less than or equal to". `(lte a b)` returns `%.y` if `a` is less than `b`, and `%.n` otherwise.
54`lte` is the standard library function for "less than or equal to". `(lte a b)` returns `%.y` if `a` is less than `b`, and `%.n` otherwise.
56+The last boolean test we have is
71−This terminates the list of children of `?&`.
60+This checks to see if `0` is equal to `(mod n 2)` and produces `%.y` if `n` is even and `%.n` if `n` is odd.
74−`?:` (pronounced "wut-pam") is the simplest "wut" rune. It takes three children, also called sub-expressions. The first child is a boolean test, which means that it looks for a `%.y` ("yes") or a `%.n` ("no."). The second child is a yes-branch, which is what we arrive at if the aforementioned boolean test evaluates to `%.y`. The third child is a no-branch, so we arrive at it if instead the boolean test evaluates to `%.n`. These branches can contain any sort of Hoon expression, including further conditional expressions, as we will see.
76−In our case, the first child of `?:` is `?& =(0 (mod n 2))`. It itself has another conditional rune, `?&` ("wut-pam"), which performs the logical "and" operation on its two children, making sure that both of them are true. The first child of our `?&` rune is `=(0 (mod n 2))`, which simply asks if `n` is even or not.
62+It is good practice in Hoon to put "lighter" lines at the top and "heavier" lines at the bottom, which is why we have put `=(0 (mod n 2))` last in the list of conditionals.
64+Finally we have a terminator.
85−
68+This marks the end of the list of children of `?&`.
86−Its first child is another `?&` rune. This second "and" rune checks for the truth of the following two expressions: `(gte n 1)`, meaning "n is greater than or equal to one"; and `?! (gth n 100),` which means "n is not greater than 100".
88−`?!` ("wut-zup") is the logical "not" operator, which inverts the truth value of its single child. We would normally use simpler code here to do the same thing: `(lte n 100)`, without the `?!` rune. We're just using this rune artificially to demonstrate its use to the reader. The `==` runes close the `?&` runes, since `?&` runes can take an unlimited number of children.
90−So the two `?&` runes that we used in this code together check the that all three of these different expressions evaluate to true: `=(0 (mod n 2))`, `(gte n 1)`, and `?! (gth n 100)`. If all of these expressions evaluate to true, our original `?:` expression branches to the `%.y` we wrote on the second-to-last line; if one or more evaluates to false, it branches to the `%.n` on the final line. Try to visualize how this works by looking through the program and examining the indentation.
92### Other Runes
70### Other Runes
@@ -96,10 +74,9 @@ We only utilized one "wut" rune in our walkthrough, but there are many others. H
96- `?:` (pronounced "wut-pam") is the simplest "wut" rune. It takes three children, also called sub-expressions. The first child is a boolean test, so it looks for a `%.y` or a `%.n`. The second child is a yes-branch, which is what we arrive at if the aforementioned boolean test evaluates to `%.y`. The third child is a no-branch, which we arrive at if the boolean test evaluates to `%.n`. These branches can contain any sort of Hoon expression, including further conditional expressions. Instead of the `?&` expression in our `%say` generator above, we could have written
74- `?:` (pronounced "wut-pam") is the simplest "wut" rune. It takes three children, also called sub-expressions. The first child is a boolean test, so it looks for a `%.y` or a `%.n`. The second child is a yes-branch, which is what we arrive at if the aforementioned boolean test evaluates to `%.y`. The third child is a no-branch, which we arrive at if the boolean test evaluates to `%.n`. These branches can contain any sort of Hoon expression, including further conditional expressions. Instead of the `?&` expression in our `%say` generator above, we could have written
101− (gte n 1)
77+?: ?& (gte n 1)
102 (lte n 100)
78 (lte n 100)