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Asked: 2026-06-01T23:29:14+00:00

I’m currently fiddling around with DCPU-16 assembler (see http://0x10c.com/doc/dcpu-16.txt and http://jazzychad.net/dcpu.html ). There is

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I’m currently fiddling around with DCPU-16 assembler (see http://0x10c.com/doc/dcpu-16.txt and http://jazzychad.net/dcpu.html).

There is one thing I don’t understand in the way the assembler instructions are transformed to hex/binary.

As an example, take an instruction like

SET B, 0x0002

which is supposed to set the value of register B to decimal 2 (or hex 0x0002 or binary 0b000010)

Instruction format for DCPU-16 is

bbbbbbaaaaaaoooo

thus, 4 bits for the opcode at the lower end, 6 bits for the first value, 6 bits for the second value.

When transforming the instruction by hand, this is how I would do it:

SET == 0x1    == 0b0001
  B == 0x01   == 0b000001
       0x0002 == 0b000010

ending up with the complete instruction being

0b0000100000010001 == 0x811

but the correct value for DCPU-16 is

0b1000100000010001 == 0x8811

that is, a leading 1 is added – why is that?

I’m totally new to assembler and any other kind of hardcore low level machine instruction stuff, so please bear with me if this is a very stupid question.

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1 Answer

  1. Editorial Team

    According to the specs,

    Values: (6 bits)
    0x00-0x07: register (A, B, C, X, Y, Z, I or J, in that order)
    ...
    0x20-0x3f: literal value 0x00-0x1f (literal)

    Thus, literals 0x00-0x1f are specified by the instruction-values 0x20-0x3f – that is, the most significant-bit (out of the 6) is set. So the literal 0x02 would have the instruction-value 0x22.

    The instruction-value 0x02 refers to the C-register, so what you thought the assembled instruction should be, 0b0000100000010001 == 0x811, would actually be the instruction SET B, C.

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