计算列表中负数的出现次数

Question

I'm trying to learn LC-3 assembly and am looking at the below example:

        .orig   x3100   
        ADD R3, R0, #0  ;copy R0 into R3
        AND R1, R1, #0  ;clear count

        ADD R3, R3, #0  ;test for Neg
        BRZP NEXT       ;count if Neg
        ADD R1, R1, #1

NEXT    AND R2, R2, #0  ;check remaining 15 bits
        ADD R2, R2, #-15
LOOP    ADD R3, R3, R3  ;shift R3 left
        BRZP AGAIN      ;count if Neg
        ADD R1, R1, #1
AGAIN   ADD R2, R2, #1  ;loop until done
        BRN LOOP

There's a couple points I don't quite understand:

ADD R3, R3, #0 ;test for Neg

I don't see how this is testing for a negative value: I'm reading it as it adds nothing to R3, ie it does nothing. The following line as well, I don't quite understand what it's doing with BRZP .

I want to change this example to instead check from a list of integers:

INTEGERS    .fill    84
            .fill    -2
            .fill    -13
            .fill    4
            .fill    -4

In the above, there's three negative integers, so the count (R1) would then be 3 at program end. How would I do this?

Answer 1

I assume ADD R3, R3, #0 ;test for Neg sets flags based on the result, so the conditional branch will be based on the result that ADD stores into R3 .

As Jester points out, it seems to be looping on counting set bits in a single integer, using add r3,r3,r3 to left-shift it and set flags.

---

I don't know lc3 specifically, but it looks like you could save an instruction by clearing the counter and having the ADD R3, R0, #0 ;copy R0 into R3 take care of setting flags initially.

LC3 doesn't have mov instructions, I take it. You copy data around by adding an immediate zero, since unlike x86, it uses 3-operand instructions where the dest doesn't have to be one of the src instructions.