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