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@JayHelton
Last active December 30, 2020 19:04
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chip 8 emu exercise
struct CPU {
registers: [u8; 16],
position_in_memory: usize,
memory: [u8; 4096],
stack: [u16; 16],
stack_pointer: usize,
}
impl CPU {
fn run(&mut self) {
loop {
let op_byte1 = self.memory[self.position_in_memory] as u16;
let op_byte2 = self.memory[self.position_in_memory + 1] as u16;
let opcode = op_byte1 << 8 | op_byte2;
let x = ((opcode & 0x0F00) >> 8) as u8;
let y = ((opcode & 0x00F0) >> 4) as u8;
let op_minor = (opcode & 0x000F) as u8;
let addr = opcode & 0x0FFF;
self.position_in_memory += 2;
println!("{:04x}", opcode);
match opcode {
0x0000 => {
println!("EXIT");
return;
}
0x00EE => {
println!("RETURN");
self.ret();
}
0x2000...0x2FFF => {
println!("CALL {:04x}", addr);
self.call(addr);
}
0x8000...0x8FFF => match op_minor {
4 => {
println!("ADD positions, {} {}", x, y);
self.add_xy(x, y);
}
_ => {
unimplemented!("opcode: {:04x}", opcode);
}
},
_ => unimplemented!("opcode {:04x}", opcode),
}
}
}
// when a function is called, we need to add the address to the instructions on the stack
// when we return from a function, we remove that address and go back to the other address
fn call(&mut self, addr: u16) {
let sp = self.stack_pointer;
let stack = &mut self.stack;
if sp > stack.len() {
panic!("Stack overflow!")
}
println!("Previous address {:04x}", self.position_in_memory);
stack[sp] = self.position_in_memory as u16;
self.stack_pointer += 1;
println!("{:04x} calling this address", addr);
self.position_in_memory = addr as usize;
}
fn ret(&mut self) {
if self.stack_pointer == 0 {
panic!("Stack underflow");
}
self.stack_pointer -= 1;
println!(
"{:04x} returning to this address",
self.stack[self.stack_pointer]
);
self.position_in_memory = self.stack[self.stack_pointer] as usize;
}
fn add_xy(&mut self, x: u8, y: u8) {
self.registers[x as usize] += self.registers[y as usize];
}
}
fn main() {
let mut cpu = CPU {
registers: [0; 16],
memory: [0; 4096],
position_in_memory: 0,
stack: [0; 16],
stack_pointer: 0,
};
cpu.registers[0] = 5;
cpu.registers[1] = 10;
cpu.memory[0x000] = 0x21; // 1) CALL instruction at 0x100
cpu.memory[0x001] = 0x00;
cpu.memory[0x002] = 0x21; // 4) CALL instruction at 0x100 again
cpu.memory[0x003] = 0x00;
cpu.memory[0x100] = 0x80; // 2 ) Add registers 0 and 1 together, then put the result in register 0, increment to next full instrauction in mem
cpu.memory[0x101] = 0x14; // (part of instruction 2)
cpu.memory[0x102] = 0x80; // 3 ) Add registers 0 and 1 together, then put the result in register 0, increment to next full instrauction in mem
cpu.memory[0x103] = 0x14; // (part of instruction 3)
cpu.memory[0x104] = 0x00; // 4 ) RETURN from current stack reference, deincrement stack, go to previous address ()
cpu.memory[0x105] = 0xEE; // (part of instruction 4)
cpu.run();
assert_eq!(cpu.registers[0], 45);
println!("5 + (10 * 2) + (10 * 2) = {}", cpu.registers[0]);
}
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