Although the words mentioned in the previous paragraph are occasionally useful (especially the simpler dup, drop and swap ), you should write code that does as little stack shuffling as possible. This requires practice getting the function arguments in the right order. Nevertheless, there are certain common patterns of needed stack manipulation that are better abstracted away into their own words.

Suppose we want to define a word to determine whether a given number n is prime. A simple algorithm is to test each number from 2 to the square root of n and see whether it is a divisor of n. In this case, n is used in two places: as an upper bound for the sequence, and as the number to test for divisibility.

The word bi applies two different quotations to the single element on the stack above them, and this is precisely what we need. For instance 5 [ 2 * ] [ 3 + ] bi yields

bi applies the quotation [ 2 * ] to the value 5 and then the quotation [ 3 + ] to the value 5 leaving us with 10 and then 8 on the stack. Without bi, we would have to first dup 5, then multiply, and then swap the result of the multiplication with the second 5, so we could do the addition

You can see that bi replaces a common pattern of dup, then calculate, then swap and calculate again.

To continue our prime example, we need a way to make a range starting from 2. We can define our own word for this [2..b], using the [a..b] word:

What's up with that inline word? This is one of the modifiers we can use after defining a word, another one being recursive. This will allow us to have the definition of a short word inlined wherever it is used, rather than incurring a function call.

Try our new [2..b] word and see that it works:

Using [2..b] to produce the range of numbers from 2 to the square root of an n that is already on the stack is easy: sqrt floor [2..b] (technically floor isn't necessary here, as [a..b] works for non-integer bounds). Let's try that out

Now, we need a word to test for divisibility. A quick search in the online help shows that divisor? is the word we want. It will help to have the arguments for testing divisibility in the other direction, so we define multiple?:

We can verify that both of these return t.

Since we're going to use bi in our prime? definition, we need a second quotation. Our second quotation needs to test for a value in the range being a divisor of n - in other words we need to partially apply the word multiple?. This can be done with the word curry, like this: [ multiple? ] curry.

Finally, once we have the range of potential divisors and the test function on the stack, we can test whether any element satisfied divisibility with any? and then negate that answer with not. Our full definition of prime looks like

Altough the definition of prime is complicated, the stack shuffling is minimal and is only used in the small helper functions, which are simpler to reason about than prime?.

Notice that prime? uses two levels of quotation nesting since bi operates on two quotations, and our second quotation contains the word curry, which also operates on a quotation. In general, Factor words tend to be rather shallow, using one level of nesting for each higher-order function, unlike Lisps or more generally languages based on the lambda calculus, which use one level of nesting for each function, higher-order or not. bi and its relative tri are a small subset of the shuffle words you will use in Factor. You ahouls also become familiar with bi, tri, and bi@ by reading about them in the online help and trying them out in the listener.

Suppose we want to define a word to determine whether a given number n is prime. A simple algorithm is to test each number from 2 to the square root of n and see whether it is a divisor of n. In this case, n is used in two places: as an upper bound for the sequence, and as the number to test for divisibility.

The word bi applies two different quotations to the single element on the stack above them, and this is precisely what we need. For instance 5 [ 2 * ] [ 3 + ] bi yields

10
8

bi applies the quotation [ 2 * ] to the value 5 and then the quotation [ 3 + ] to the value 5 leaving us with 10 and then 8 on the stack. Without bi, we would have to first dup 5, then multiply, and then swap the result of the multiplication with the second 5, so we could do the addition

5 dup 2 * swap 3 +

You can see that bi replaces a common pattern of dup, then calculate, then swap and calculate again.

To continue our prime example, we need a way to make a range starting from 2. We can define our own word for this [2..b], using the [a..b] word:

: [2..b] ( n -- {2,...,n} ) 2 swap [a..b] ; inline

What's up with that inline word? This is one of the modifiers we can use after defining a word, another one being recursive. This will allow us to have the definition of a short word inlined wherever it is used, rather than incurring a function call.

Try our new [2..b] word and see that it works:

6 [2..b] >array .

Using [2..b] to produce the range of numbers from 2 to the square root of an n that is already on the stack is easy: sqrt floor [2..b] (technically floor isn't necessary here, as [a..b] works for non-integer bounds). Let's try that out

16 sqrt [2..b] >array .

Now, we need a word to test for divisibility. A quick search in the online help shows that divisor? is the word we want. It will help to have the arguments for testing divisibility in the other direction, so we define multiple?:

: multiple? ( a b -- ? ) swap divisor? ; inline

We can verify that both of these return t.

9 3 divisor? .
3 9 multiple? .

Since we're going to use bi in our prime? definition, we need a second quotation. Our second quotation needs to test for a value in the range being a divisor of n - in other words we need to partially apply the word multiple?. This can be done with the word curry, like this: [ multiple? ] curry.

Finally, once we have the range of potential divisors and the test function on the stack, we can test whether any element satisfied divisibility with any? and then negate that answer with not. Our full definition of prime looks like

: prime? ( n -- ? )
[ sqrt [2..b] ] [ [ multiple? ] curry ] bi any? not ;

Altough the definition of prime is complicated, the stack shuffling is minimal and is only used in the small helper functions, which are simpler to reason about than prime?.

Notice that prime? uses two levels of quotation nesting since bi operates on two quotations, and our second quotation contains the word curry, which also operates on a quotation. In general, Factor words tend to be rather shallow, using one level of nesting for each higher-order function, unlike Lisps or more generally languages based on the lambda calculus, which use one level of nesting for each function, higher-order or not. bi and its relative tri are a small subset of the shuffle words you will use in Factor. You ahouls also become familiar with bi, tri, and bi@ by reading about them in the online help and trying them out in the listener.

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Factor 0.99 x86.64 (2190, heads/master-46c625f8dc, Feb 4 2023 19:37:27)