80 lines
3.1 KiB
Plaintext
80 lines
3.1 KiB
Plaintext
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== Synacor Challenge ==
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In this challenge, your job is to use this architecture spec to create a
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virtual machine capable of running the included binary. Along the way,
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you will find codes; submit these to the challenge website to track
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your progress. Good luck!
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== architecture ==
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- three storage regions
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- memory with 15-bit address space storing 16-bit values
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- eight registers
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- an unbounded stack which holds individual 16-bit values
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- all numbers are unsigned integers 0..32767 (15-bit)
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- all math is modulo 32768; 32758 + 15 => 5
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== binary format ==
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- each number is stored as a 16-bit little-endian pair (low byte, high byte)
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- numbers 0..32767 mean a literal value
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- numbers 32768..32775 instead mean registers 0..7
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- numbers 32776..65535 are invalid
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- programs are loaded into memory starting at address 0
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- address 0 is the first 16-bit value, address 1 is the second 16-bit value, etc
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== execution ==
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- 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.
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- Encountering a register as an operation argument should be taken as reading from the register or setting into the register as appropriate.
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== hints ==
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- Start with operations 0, 19, and 21.
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- Here's a code for the challenge website: ............
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- The program "9,32768,32769,4,19,32768" occupies six memory addresses and should:
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- Store into register 0 the sum of 4 and the value contained in register 1.
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- Output to the terminal the character with the ascii code contained in register 0.
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== opcode listing ==
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halt: 0
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stop execution and terminate the program
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set: 1 a b
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set register <a> to the value of <b>
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push: 2 a
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push <a> onto the stack
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pop: 3 a
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remove the top element from the stack and write it into <a>; empty stack = error
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eq: 4 a b c
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set <a> to 1 if <b> is equal to <c>; set it to 0 otherwise
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gt: 5 a b c
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set <a> to 1 if <b> is greater than <c>; set it to 0 otherwise
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jmp: 6 a
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jump to <a>
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jt: 7 a b
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if <a> is nonzero, jump to <b>
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jf: 8 a b
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if <a> is zero, jump to <b>
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add: 9 a b c
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assign into <a> the sum of <b> and <c> (modulo 32768)
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mult: 10 a b c
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store into <a> the product of <b> and <c> (modulo 32768)
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mod: 11 a b c
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store into <a> the remainder of <b> divided by <c>
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and: 12 a b c
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stores into <a> the bitwise and of <b> and <c>
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or: 13 a b c
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stores into <a> the bitwise or of <b> and <c>
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not: 14 a b
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stores 15-bit bitwise inverse of <b> in <a>
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rmem: 15 a b
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read memory at address <b> and write it to <a>
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wmem: 16 a b
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write the value from <b> into memory at address <a>
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call: 17 a
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write the address of the next instruction to the stack and jump to <a>
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ret: 18
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remove the top element from the stack and jump to it; empty stack = halt
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out: 19 a
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write the character represented by ascii code <a> to the terminal
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in: 20 a
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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
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noop: 21
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no operation
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