**i= 0 then x[i]= even, i =1 then [i]= odd** convert from assembly to Java programming

Question

Here is the question from assessed exercise as below. Even though I found and summarize the answer from the web, I still don't figure out how it works in Java programming which can assist a fresh learner to understand the logic and hence I am more likely to convert it to assembly language in Sigma 16 system.

Q: A Sigma16 system has an array, X , of 16-bit signed numbers in memory. Write an assembly language program to write a second array, Y whose ith element is 0 if the ith element of X is even and 1 if it is odd.

Ans:

        ADD R1, R0, R0  ; i = 0
        LEA R2, 4[R0]   ; R2 = 4
        LEA R3, 1[R0]   ; R3 = 1
        LEA R4, 2[R0]   ; R4 = 2
Loop    LOAD R5, X[R1]  ; R5 = x[i]

        DIV R6, R5, R4      ; x[i] / 2
        STORE R15, Y[R1]    ; y[i] = x[i] mod 2

        ADD R1, R1, R3      ; i++
        CMPLT R15, R1, R2   ; If i< 4 then⋯
        JUMPT R15, Loop[R0] ; Loop

        TRAP R0,R0,R0

Your answers are highly appreciated. Please also correct the above codes if they are way biased with errors.

Answer 1

16-bit signed integers in Java are short s, so an equivalent code fragment in Java might be:

short[]x = new short[]{32, 203, 98, 19, 47, 78}; 
short[]y = new short[x.length]; 

for (int i = 0; i < x.length; ++x)
    if (x[i] % 2 == 0)  // % is the remainder operator 
        y[i] = 0;
    else
        y[i] = 1;  

You could also do the odd-even test in one line with

    y[i] = (x[i] % 2 == 0) ? 0 : 1;

And rather than a "remainder-after-division-by-2" test, there is also the option of a more assembly-like bitwise operation:

    y[i] = x[i] & 1;