== 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