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assemblyemulationbit-manipulationintel-8080

Intel 8080: Memory offset calculation

I am reading this guide about Intel 8080 emulation Emulator 101 and when I'm reading the code to check what I wrote, I stumbled upon this

case 0x36:      //MVI   M,byte
{                   
//AC set if lower nibble of h was zero prior to dec
uint16_t offset = (state->h<<8) | state->l;
state->memory[offset] = opcode[1];
state->pc++;
}
break;

from a book called Intel 8080/8085 Assembly Language Programming, I read about MVI this

This instruction copies the data stored in its second byte into the memory location addressed by H and L. M is a symbolic reference to the H and L register pair.

so I'm guessing that the offset is the memory location addressed by H and L, but why do we do it that way? That is (state->h<<8) | state->l

Thanks

Solution

  • how is the combination of << and | gives us the 16-bit offset?

    Take H, an 8-bit register, where H7 is the most significant bit and H0 is the least significant bit:

    H7H6H5H4H3H2H1H0

    Take L, an 8-bit register, where L7 is the most significant bit and L0 is the least significant bit: 

    L7L6L5L4L3L2L1L0

    You now want to construct the 16-bit offset that results from combining H (Highest 8-bits) and L (Lowest 8-bits.) In C/C++/Java this can be achieved by an 8-bit shift-left << followed by a bitwise-or | as follows:

    H      =                 H7H6H5H4H3H2H1H0
H<<8   = H7H6H5H4H3H2H1H00 0 0 0 0 0 0 0 
H<<8|L = H7H6H5H4H3H2H1H0L7L6L5L4L3L2L1L0