Search code examples
assemblycpu-registers8051accumulator

Accumulator register in 8051 microprocessor

We're tasked to identify the problem in next code:

ORG 0000H
MAIN:
    CLR A
    MOV A, FFFFH
    MOV B, FFH
AGAIN: DEC A
    DJNZ B, AGAIN
END

On MOV A, FFFFH and MOV B, FFH EdSim51 is having an

unknow label on both FFFFH and FFH.

I don't have any experience on this type of code. I think it's on the amount of bits each register holds, but I'm not sure. Can someone explain it to me.

Solution

  • The instruction MOV can be used with different operands. In your case, the assembler looks for an "immediate value". This could be given as a label or a literal number.

    For us humans, both FFFFH and FFH look as literal numbers because we recognize the hexadecimal digits and the "H" at the end.

    But the software uses a most simple method: If the first character is a decimal digit, the operand is a literal number. If it is a letter, it is a label.

    That's why EdSim51 (or its assembler, respectively) takes them as labels. And since none of the labels is defined, it reports the error you get.

    However, if you put a zero "0" in front of both, resulting in 0FFFFH and 0FFH, the first instruction might give you another error: The value is too large for the instruction, which expects an 8-bit width.