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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, ...)