DCPU-16 Specification
Copyright 2012 Mojang
Version 1.1 (Check 0x10c.com for updated versions)
* 16 bit unsigned words
* 0x10000 words of ram
* 8 registers (A, B, C, X, Y, Z, I, J)
* program counter (PC)
* stack pointer (SP)
* overflow (O)
In this document, anything within [brackets] is shorthand for "the value of the RAM at the location of the value inside the brackets".
For example, SP means stack pointer, but [SP] means the value of the RAM at the location the stack pointer is pointing at.
Whenever the CPU needs to read a word, it reads [PC], then increases PC by one. Shorthand for this is [PC++].
In some cases, the CPU will modify a value before reading it, in this case the shorthand is [++PC].
Instructions are 1-3 words long and are fully defined by the first word.
In a basic instruction, the lower four bits of the first word of the instruction are the opcode,
and the remaining twelve bits are split into two six bit values, called a and b.
a is always handled by the processor before b, and is the lower six bits.
In bits (with the least significant being last), a basic instruction has the format: bbbbbbaaaaaaoooo
Values: (6 bits)
0x00-0x07: register (A, B, C, X, Y, Z, I or J, in that order)
0x08-0x0f: [register]
0x10-0x17: [next word + register]
0x18: POP / [SP++]
0x19: PEEK / [SP]
0x1a: PUSH / [--SP]
0x1b: SP
0x1c: PC
0x1d: O
0x1e: [next word]
0x1f: next word (literal)
0x20-0x3f: literal value 0x00-0x1f (literal)
* "next word" really means "[PC++]". These increase the word length of the instruction by 1.
* If any instruction tries to assign a literal value, the assignment fails silently. Other than that, the instruction behaves as normal.
* All values that read a word (0x10-0x17, 0x1e, and 0x1f) take 1 cycle to look up. The rest take 0 cycles.
* By using 0x18, 0x19, 0x1a as POP, PEEK and PUSH, there's a reverse stack starting at memory location 0xffff. Example: "SET PUSH, 10", "SET X, POP"
Basic opcodes: (4 bits)
0x0: non-basic instruction - see below
0x1: SET a, b - sets a to b
0x2: ADD a, b - sets a to a+b, sets O to 0x0001 if there's an overflow, 0x0 otherwise
0x3: SUB a, b - sets a to a-b, sets O to 0xffff if there's an underflow, 0x0 otherwise
0x4: MUL a, b - sets a to a*b, sets O to ((a*b)>>16)&0xffff
0x5: DIV a, b - sets a to a/b, sets O to ((a<<16)/b)&0xffff. if b==0, sets a and O to 0 instead.
0x6: MOD a, b - sets a to a%b. if b==0, sets a to 0 instead.
0x7: SHL a, b - sets a to a<**>16)&0xffff
0x8: SHR a, b - sets a to a>>b, sets O to ((a<<16)>>b)&0xffff
0x9: AND a, b - sets a to a&b
0xa: BOR a, b - sets a to a|b
0xb: XOR a, b - sets a to a^b
0xc: IFE a, b - performs next instruction only if a==b
0xd: IFN a, b - performs next instruction only if a!=b
0xe: IFG a, b - performs next instruction only if a>b
0xf: IFB a, b - performs next instruction only if (a&b)!=0
* SET, AND, BOR and XOR take 1 cycle, plus the cost of a and b
* ADD, SUB, MUL, SHR, and SHL take 2 cycles, plus the cost of a and b
* DIV and MOD take 3 cycles, plus the cost of a and b
* IFE, IFN, IFG, IFB take 2 cycles, plus the cost of a and b, plus 1 if the test fails
Non-basic opcodes always have their lower four bits unset, have one value and a six bit opcode.
In binary, they have the format: aaaaaaoooooo0000
The value (a) is in the same six bit format as defined earlier.
Non-basic opcodes: (6 bits)
0x00: reserved for future expansion
0x01: JSR a - pushes the address of the next instruction to the stack, then sets PC to a
0x02-0x3f: reserved
* JSR takes 2 cycles, plus the cost of a.
FAQ:
Q: Why is there no JMP or RET?
A: They're not needed! "SET PC, " is a one-instruction JMP.
For small relative jumps in a single word, you can even do "ADD PC, " or "SUB PC, ".
For RET, simply do "SET PC, POP"
Q: How does the overflow (O) work?
A: O is set by certain instructions (see above), but never automatically read. You can use its value in instructions, however.
For example, to do a 32 bit add of 0x12345678 and 0xaabbccdd, do this:
SET [0x1000], 0x5678 ; low word
SET [0x1001], 0x1234 ; high word
ADD [0x1000], 0xccdd ; add low words, sets O to either 0 or 1 (in this case 1)
ADD [0x1001], O ; add O to the high word
ADD [0x1001], 0xaabb ; add high words, sets O again (to 0, as 0xaabb+0x1235 is lower than 0x10000)
Q: How do I do 32 or 64 bit division using O?
A: This is left as an exercise for the reader.
Q: How about a quick example?
A: Sure! Here's some sample assembler, and a memory dump of the compiled code:
; Try some basic stuff
SET A, 0x30 ; 7c01 0030
SET [0x1000], 0x20 ; 7de1 1000 0020
SUB A, [0x1000] ; 7803 1000
IFN A, 0x10 ; c00d
SET PC, crash ; 7dc1 001a [*]
; Do a loopy thing
SET I, 10 ; a861
SET A, 0x2000 ; 7c01 2000
:loop SET [0x2000+I], [A] ; 2161 2000
SUB I, 1 ; 8463
IFN I, 0 ; 806d
SET PC, loop ; 7dc1 000d [*]
; Call a subroutine
SET X, 0x4 ; 9031
JSR testsub ; 7c10 0018 [*]
SET PC, crash ; 7dc1 001a [*]
:testsub SHL X, 4 ; 9037
SET PC, POP ; 61c1
; Hang forever. X should now be 0x40 if everything went right.
:crash SET PC, crash ; 7dc1 001a [*]
; [*]: Note that these can be one word shorter and one cycle faster by using the short form (0x00-0x1f) of literals,
; but my assembler doesn't support short form labels yet.
Full memory dump:
0000: 7c01 0030 7de1 1000 0020 7803 1000 c00d
0008: 7dc1 001a a861 7c01 2000 2161 2000 8463
0010: 806d 7dc1 000d 9031 7c10 0018 7dc1 001a
0018: 9037 61c1 7dc1 001a 0000 0000 0000 0000
**