I’m trying to learn redcode, because it looks fun to make a bot.
Introduction
For those who don’t know what redcode is, here’s a short explanation. It’s an ASM-like language, but far more easy and stripped. It is used to write little programs that need to shut down other programs in virtual memory. (See for more info here: http://vyznev.net/corewar/guide.html)
Here’s a piece of code:
;redcode ;name Mice ;author Chip Wendell ;strategy paper (replicator) ;history Winner of the 1986 ICWS tournament Top dat #0, #0 Start mov #12, Top Loop mov @Top, <Target djn Loop, Top spl @Target,0 Spacer equ 653 add #Spacer,Target jmz Start, Top Target dat #0, #833 end Start Problem
The basic strategy is to replicate itself to another place, and the fork the process. What I don’t understand is this rule:
Loop mov @Top, <Target I understand the meaning of this line. It says, move the B-Field of target to the line where the B-Field of top points, and decrease the value of the B-Field of target.
When loop is executed for the first time, the first line will be:
Top dat #0, #12 As far as I get, the line with Loop means: Move the instruction 12 lines ahead (filled with dat #0, #0) to line 833.
But when this code is executed, the line of code is placed at line 839.
Does someone understand what is happening really?
Okay, that took quite a bit of reading, but here’s your answer:
The first instruction, as you correctly surmise, makes Top into DAT #0, #12. Simple enough, but the next instruction is trickier. First, it decrements the B-value of Target (making it 832). Then, it copies the instruction at Top into the location 832 lines ahead relative to Target. That’s the key: the indirect addressing mode means the destination is relative to the B-value read for the offset. Now look at the code after compilation, with line numbers in front:
As you can see, Target is on line 7, so 832 lines ahead relative to Target is line 839.
Hope that clears it up for you.