Question

I need to show all numbers from matrix(3×3) where number has two first bits set.
I think i’m somewhere near the solution, but something is wrong, can you point what is wrong?

.model small
.stack 100h
.data
n equ 3
a dw n*n dup(?)


.code
 extrn write:near 
 extrn read:near

begin:
    mov ax,@data
    mov ds,ax
        mov cx,n*n 
    mov si,0   
m1: 
    call read
    mov a[si],ax  
    add si,2
    loop m1             

    mov si,0        
    mov cx,n*n
m2: 
    mov bx, a[si]
    test bx,1
    jz net
    test bx,2
    jz net
    mov ax,a[si]
    call write
net:
    add si,2
    loop m2

    mov ah,4ch
    int 21h
end begin

Answer 1

It depends on what you mean by “two first bits set.” The code you’ve written works fine if you’re looking for the two lowest-order bits. That is, bit 0 and bit 1 are set. The example you give in the comments (3, 6, 7), probably outputs 3 and 7 because they have their lowest-order two bits set. That is:

3 = 00000011 binary
6 = 00000110 binary
7 = 00000111 binary

So only 3 and 7 meet the conditions you’ve written in your code.

You appear to be interpreting “two first bits” to mean that the number starts with the binary sequence “11” after any leading zeros. If that’s really the interpretation you want, then you’ll have to use the left shift instruction until the high bit is set to 1, and then check if the next-to-highest bit is set to 1.