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Implicit Compilation in zeptoforth

tabemann edited this page Feb 3, 2024 · 2 revisions

Note: this is a new feature that is first available in release 1.5.0.

Implicit compilation is a feature of zeptoforth that allows the evaluation of code from the REPL through compiling it in RAM as a temporary anonymous word which is then executed, and afterwards immediately forgotten. It is initiated by evaluating any one of if, begin, do, ?do, or case while in interpretation mode, and execution is triggered by compiling a matching then, for if; a matching end, until, repeat (if while had been specified), or again for begin; a matching loop or +loop for do or ?do, or a matching endcase or endcasestr for case. Once it has been started no code aside from immediate words is evaluated until the outermost block is matched. All operations permissible during ordinary compilation are permissible during implicit compilation except for ones which require space alloted in the RAM dictionary to remain valid afterwards.

Examples of implicit compilation in action include:

begin 2,0 3,0 4,0 5,0 { D: a D: b D: c D: x } a x 2 fi** f* b x f* d+ c d+ f. end 69,0  ok
true if false if ." *A*" else ." *B*" then else ." *C*" then *B* ok
10 0 do i . loop 0 1 2 3 4 5 6 7 8 9  ok
10 begin ?dup while rng::random . 1- repeat -483139579 65935751 618859148 -1719283164 1699951016 1297462867 894730864 1042467505 831079856 56931937  ok
begin ." *" 250 ms key? until ******************** ok
1 case 0 of ." foo" endof 1 of ." bar" endof 2 of ." baz" endof endcase bar ok

Note that it is not safe to carry out operations which allot RAM dictionary space which is meant to persist outside of the implicitly compiled block, because all the RAM dictionary space alloted from the point that implicit compilation is initiated is reclaimed afterwards and thus is very likely to be overwritten in the future. E.g. the following is unsafe:

begin 0 [: begin ." *" 1000 ms again ;] 320 128 512 task::spawn task::run end

This is because the lambda block defined during implicit compilation will be referenced past its conclusion as it will be executing as part of a persistent task, and thus will be most likely corrupted by any future usage of the main task's RAM dictionary.

For this reason [: does not start implicit compilation by itself, because the lambda defined by it would immediately go out of scope with the matching ;]. If one wishes to use [: during implicit compilation, one must contain it in some other block, such as a begin ... end block to give the lambda so defined a scope in which it is valid.

Defining lambdas and generally alloting space in the RAM dictionary during implicit compilation is safe as long as they do not persist outside its scope, as in the following example:

begin 256 [: { addr } 256 0 do i addr i + c! loop addr addr 256 + dump ;] with-allot end 
200192A0  00 01 02 03  04 05 06 07  08 09 0A 0B  0C 0D 0E 0F  |................|
200192B0  10 11 12 13  14 15 16 17  18 19 1A 1B  1C 1D 1E 1F  |................|
200192C0  20 21 22 23  24 25 26 27  28 29 2A 2B  2C 2D 2E 2F  | !"#$%&'()*+,-./|
200192D0  30 31 32 33  34 35 36 37  38 39 3A 3B  3C 3D 3E 3F  |0123456789:;<=>?|
200192E0  40 41 42 43  44 45 46 47  48 49 4A 4B  4C 4D 4E 4F  |@ABCDEFGHIJKLMNO|
200192F0  50 51 52 53  54 55 56 57  58 59 5A 5B  5C 5D 5E 5F  |PQRSTUVWXYZ[\]^_|
20019300  60 61 62 63  64 65 66 67  68 69 6A 6B  6C 6D 6E 6F  |`abcdefghijklmno|
20019310  70 71 72 73  74 75 76 77  78 79 7A 7B  7C 7D 7E 7F  |pqrstuvwxyz{|}~.|
20019320  80 81 82 83  84 85 86 87  88 89 8A 8B  8C 8D 8E 8F  |................|
20019330  90 91 92 93  94 95 96 97  98 99 9A 9B  9C 9D 9E 9F  |................|
20019340  A0 A1 A2 A3  A4 A5 A6 A7  A8 A9 AA AB  AC AD AE AF  |................|
20019350  B0 B1 B2 B3  B4 B5 B6 B7  B8 B9 BA BB  BC BD BE BF  |................|
20019360  C0 C1 C2 C3  C4 C5 C6 C7  C8 C9 CA CB  CC CD CE CF  |................|
20019370  D0 D1 D2 D3  D4 D5 D6 D7  D8 D9 DA DB  DC DD DE DF  |................|
20019380  E0 E1 E2 E3  E4 E5 E6 E7  E8 E9 EA EB  EC ED EE EF  |................|
20019390  F0 F1 F2 F3  F4 F5 F6 F7  F8 F9 FA FB  FC FD FE FF  |................|
 ok