Lesson 19 · Memory through HL - Build a Game Boy Emulator

Memory through HL

Today you give the CPU its first way to touch memory, implementing LD (HL), A and LD A, (HL), which read and write a byte through the pointer register HL - the technique every future data-moving instruction builds on.

The goal

Implement LD (HL), A and LD A, (HL) so the CPU can read and write memory through HL as a pointer.

Start here - the target

TO DO

Scenario: Storing the accumulator through HL

GivenHL is 0xC000 and A is 0x42

WhenLD (HL), A (opcode 0x77) runs

Thenmemory at 0xC000 holds 0x42

Andthe step reports 8 cycles (a memory access costs an extra 4)

Andthe read direction LD A, (HL) (opcode 0x7E) loads that 0x42 back into A, also in 8 cycles

Background

Until now instructions have only touched registers. LD (HL), A (opcode 0x77) is the first that reaches into memory: the parentheses mean “indirect,” so (HL) is the byte at the address held in HL. This is register-indirect addressing, and HL is the CPU’s dedicated pointer for it.

Notice the cost: touching memory adds 4 cycles, so these run in 8 rather than 4. The reverse direction, LD A, (HL) (opcode 0x7E), reads the pointed-at byte back into A. Together they let a program stream data through memory one byte at a time - the foundation for copying tiles into video RAM, which is exactly what you will be doing by the end of the project.

Make it work

case 0x77: // LD (HL), A
    c.mem.Write(c.HL(), c.A)
    return 8
case 0x7E: // LD A, (HL)
    c.A = c.mem.Read(c.HL())
    return 8