2.7 KiB
2.7 KiB
Bytecode
TODO
Example
Let's consider a simple algorithm that tests the Collatz conjecture (which says that applying a simple operation from any starting number over and over will always lead to number 1). The algorithm in C would look as follows:
// Collatz conjecture
#include <stdio.h>
int next(int n)
{
return n % 2 ? // is odd?
3 * n + 1 :
n / 2;
}
int main(void)
{
int n = getchar() - '0'; // read input ASCII digit
while (1)
{
printf("%d\n",n);
if (n == 1)
break;
n = next(n);
}
return 0;
}
The program reads a number (one digit for simplicity) and then prints the sequence until reaching the final number 1. Now let's rewrite the same algorithm in comun, a language which will allow us to produce bytecode:
# Collatz conjecture
next:
$0 2 % ? # is odd?
3 * 1 +
;
2 /
.
.
<- # read input ASCII digit
"0" - # convert it to number
@@
# print:
$0 10 / "0" + ->
$0 10 % "0" + ->
10 ->
$0 1 = ?
!@
.
next
.
The bytecode this compiles to is following:
000000: DES 00 0111 # func
000001: JMA 00 0100... # 20 (#14)
000002: COC 00 0001
000003: MGE 00 0000
000004: CON' 00 0010 # 2 (#2)
000005: MOX 00 0000
000006: DES 00 0001 # if
000007: JNA 00 0000... # 16 (#10)
000008: COC 00 0001
000009: CON' 00 0011 # 3 (#3)
00000a: MUX 00 0000
00000b: CON' 00 0001 # 1 (#1)
00000c: ADX 00 0000
00000d: DES 00 0010 # else
00000e: JMA 00 0011... # 19 (#13)
00000f: COC 00 0001
000010: CON' 00 0010 # 2 (#2)
000011: DIX 00 0000
000012: DES 00 0011 # end if
000013: RET 00 0000
000014: INI 00 0000
000015: INP 00 0000
000016: CON' 00 0000... # 48 (#30)
000017: COC 00 0011
000018: SUX 00 0000
000019: DES 00 0100 # loop
00001a: MGE 00 0000
00001b: CON' 00 1010 # 10 (#a)
00001c: DIX 00 0000
00001d: CON' 00 0000... # 48 (#30)
00001e: COC 00 0011
00001f: ADX 00 0000
000020: OUT 00 0000
000021: MGE 00 0000
000022: CON' 00 1010 # 10 (#a)
000023: MOX 00 0000
000024: CON' 00 0000... # 48 (#30)
000025: COC 00 0011
000026: ADX 00 0000
000027: OUT 00 0000
000028: CON' 00 1010 # 10 (#a)
000029: OUT 00 0000
00002a: MGE 00 0000
00002b: CON' 00 0001 # 1 (#1)
00002c: EQX 00 0000
00002d: DES 00 0001 # if
00002e: JNA 00 0100... # 52 (#34)
00002f: COC 00 0011
000030: DES 00 0101 # break
000031: JMA 00 1000... # 56 (#38)
000032: COC 00 0011
000033: DES 00 0011 # end if
000034: CAL 00 0011 # 3 (#3)
000035: DES 00 0110 # end loop
000036: JMA 00 1010... # 26 (#1a)
000037: COC 00 0001
000038: END 00 0000
TODO: analyze the above, show other bytecodes (python, java, ...)