Haskell implementation of the Synacor VM.
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README.md
Vmachine.hs
challenge.bin

README.md

== Synacor Challenge == In this challenge, your job is to use this architecture spec to create a virtual machine capable of running the included binary. Along the way, you will find codes; submit these to the challenge website to track your progress. Good luck!

== architecture ==

  • three storage regions
    • memory with 15-bit address space storing 16-bit values
    • eight registers
    • an unbounded stack which holds individual 16-bit values
  • all numbers are unsigned integers 0..32767 (15-bit)
  • all math is modulo 32768; 32758 + 15 => 5

== binary format ==

  • each number is stored as a 16-bit little-endian pair (low byte, high byte)
  • numbers 0..32767 mean a literal value
  • numbers 32768..32775 instead mean registers 0..7
  • numbers 32776..65535 are invalid
  • programs are loaded into memory starting at address 0
  • address 0 is the first 16-bit value, address 1 is the second 16-bit value, etc

== execution ==

  • After an operation is executed, the next instruction to read is immediately after the last argument of the current operation. If a jump was performed, the next operation is instead the exact destination of the jump.
  • Encountering a register as an operation argument should be taken as reading from the register or setting into the register as appropriate.

== hints ==

  • Start with operations 0, 19, and 21.
  • Here's a code for the challenge website: mzMXMdUpOZeH
  • The program "9,32768,32769,4,19,32768" occupies six memory addresses and should:
    • Store into register 0 the sum of 4 and the value contained in register 1.
    • Output to the terminal the character with the ascii code contained in register 0.

== opcode listing ==

  • halt: 0 stop execution and terminate the program
  • set: 1 a b set register to the value of
  • push: 2 a push onto the stack
  • pop: 3 a remove the top element from the stack and write it into ; empty stack = error
  • eq: 4 a b c set to 1 if is equal to ; set it to 0 otherwise
  • gt: 5 a b c set to 1 if is greater than ; set it to 0 otherwise
  • jmp: 6 a jump to
  • jt: 7 a b if is nonzero, jump to
  • jf: 8 a b if is zero, jump to
  • add: 9 a b c assign into the sum of and (modulo 32768)
  • mult: 10 a b c store into the product of and (modulo 32768)
  • mod: 11 a b c store into the remainder of divided by
  • and: 12 a b c stores into the bitwise and of and
  • or: 13 a b c stores into the bitwise or of and
  • not: 14 a b stores 15-bit bitwise inverse of in
  • rmem: 15 a b read memory at address and write it to
  • wmem: 16 a b write the value from into memory at address
  • call: 17 a write the address of the next instruction to the stack and jump to
  • ret: 18 remove the top element from the stack and jump to it; empty stack = halt
  • out: 19 a write the character represented by ascii code to the terminal
  • in: 20 a read a character from the terminal and write its ascii code to ; it can be assumed that once input starts, it will continue until a newline is encountered; this means that you can safely read whole lines from the keyboard and trust that they will be fully read
  • noop: 21 no operation