Initial commit: implementation of the virtual machine

arch-spec

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+== 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: SJIDydAvKFEH
+- 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 <a> to the value of <b>
+push: 2 a
+  push <a> onto the stack
+pop: 3 a
+  remove the top element from the stack and write it into <a>; empty stack = error
+eq: 4 a b c
+  set <a> to 1 if <b> is equal to <c>; set it to 0 otherwise
+gt: 5 a b c
+  set <a> to 1 if <b> is greater than <c>; set it to 0 otherwise
+jmp: 6 a
+  jump to <a>
+jt: 7 a b
+  if <a> is nonzero, jump to <b>
+jf: 8 a b
+  if <a> is zero, jump to <b>
+add: 9 a b c
+  assign into <a> the sum of <b> and <c> (modulo 32768)
+mult: 10 a b c
+  store into <a> the product of <b> and <c> (modulo 32768)
+mod: 11 a b c
+  store into <a> the remainder of <b> divided by <c>
+and: 12 a b c
+  stores into <a> the bitwise and of <b> and <c>
+or: 13 a b c
+  stores into <a> the bitwise or of <b> and <c>
+not: 14 a b
+  stores 15-bit bitwise inverse of <b> in <a>
+rmem: 15 a b
+  read memory at address <b> and write it to <a>
+wmem: 16 a b
+  write the value from <b> into memory at address <a>
+call: 17 a
+  write the address of the next instruction to the stack and jump to <a>
+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 <a> to the terminal
+in: 20 a
+  read a character from the terminal and write its ascii code to <a>; 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