Remove/refactor the descriptions.

This commit is contained in:
HorseMD 2014-11-13 16:10:56 +00:00
parent 12e22592e7
commit c0774dc821

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@ -21,31 +21,26 @@ of what is written here should work elsewhere.
\ It's important to know how forth processes instructions. All programming in Forth is
\ done by manipulating what's known as the parameter stack (more commonly just referred
\ to as "the stack"). The stack is a typical last-in-first-out (LIFO) stack. Typing:
\ to as "the stack"). Typing:
5 2 3 56 76 23 65
\ Means 5 gets put on the stack first, then 2, then 3, etc all the way to 65, which
\ is now at the top of the stack. We can see the length and contents of the stack by
\ passing forth the word `.s`:
.s <7> 5 2 3 56 76 23 65 \ ok
\ Makes those numbers get added to the stack, from left to right.
.s \ <7> 5 2 3 56 76 23 65 ok
\ Forth's interpreter interprets what you type in one of two ways: as *words* (i.e. the
\ name of subroutines) or as *numbers*. Words are essentially "symbols that do things".
\ Finally, as the stack is LIFO, we obviously must use postfix notation to manipulate
\ the stack. This should become clear shortly.
\ ------------------------------ Basic Arithmetic ------------------------------
\ Lets do a simple equation: adding 5 and 4. In infix notation this would be 5 + 4,
\ but as forth works in postfix (see above about stack manipulation) we input it like so:
\ Arithmetic (in fact most words requiring data) works by manipulating data on
\ the stack.
5 4 + \ ok
\ However, this alone yields "ok", yet no answer. Typing the word `.` will yield
\ the result.
\ This adds 5 and 4 to the stack and then `+` is called, which removes them and
\ adds the result to the stack. We can see it with `.`:
. \ 9 ok
\ This should illustrate how Forth's stack works. Lets do a few more arithmetic tests:
\ A few more examples of arithmetic
6 7 * . \ 42 ok
1360 23 - . \ 1337 ok
12 12 / . \ 1 ok
@ -75,19 +70,15 @@ of what is written here should work elsewhere.
\ Quite often one will want to write their own words.
: square ( n -- n ) dup * ; \ ok
\ Lets break this down. The `:` word says to Forth to enter "compile" mode. After that,
\ we tell Forth what our word is called - "square". Between the parentheses we have a
\ comment depicting what this word does to the stack - it takes a number and adds a
\ number. Finally, we have what the word does, until we reach the `;` word which
\ says that you've finished your definition, Forth will add this to the dictionary and
\ switch back into interpret mode.
\ The `:` word sets forth into compile mode. `(` and `)` are both words which
\ tell forth to ignore between them. Up until the `;` word is what our word
\ does.
\ We can check the definition of a word with the `see` word:
see square \ dup * ; ok
\ ------------------------------ Conditionals ------------------------------
\ Booleans:
\ In forth, -1 is used to represent truth, and 0 is used to represent false.
\ The idea is that -1 is 11111111 in binary, whereas 0 is obviously 0 in binary.
\ However, any non-zero value is usually treated as being true: