Signed overflow detection on 8080

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I want to branch if x-y overflows.

I came up with idea to store x-y into register A , store 0 into register B and compare these two registers.

Unlike 8086, 8080 doesn't have an OF flag or jo / jno instructions.

x db
y db

    lda x
    mov b,a
    lda y
    sub b
    mvi b,0
    cmp b
    jp overflow

notOverFlow HLT
overflow HLT

It works with x=128 , y=127 but not with values x=0, y=0.

intel-8080 integer-overflow assembly
1 Answers

If no overflow means the arithmetic result is representable as an 8-bit signed value, then the following rules should apply:

  • if y = 0, then no overflow.
  • if y > 0, then overflow if truncate(x - y) > x.
  • if y < 0, then overflow if truncate(x - y) < x.

Here truncate(x) means the truncated 8-bit signed value of x.

Then the code may look like this:

    lda y       ; Load y.
    mov b, a

    lda x       ; Load x.
    mov c, a

    sub b       ; No overflow if truncate(x - y) = x, that is, y = 0.
    cmp c
    jz no_overflow

    jm else     ; Jump if truncate(x - y) < x.

                ; At this point y != 0 and truncate(x - y) > x.

    mov a, b    ; Overflow if y > 0.
    ana a
    jp overflow

no_overflow:
    ...

                ; At this point y != 0 and truncate(x - y) < x.
else:
    mov a, b    ; Overflow if y < 0.
    ana a
    jp no_overflow

overflow:
    ...

As an optimization measure, the two mov a, b instructions can be replaced with a single mov a, b just before jm else.

user_114890240
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