Translating C++ to ARM Assembly using (MLA instruction)

Question

I am trying to translate this C++ code:

y = y+ x*32;
z = y+ x*x;

To ARM assembly assume (x is R1 register, y is R2, z in R3), and I should use only one assembly instruction for each case So, I suggest to do it using (MLA) but I don't know how, can you please help me!!

Answer 1

Firstly, put your code snippet in an function and create a complete code.

void x(void){
    volatile int x = 1, y = 2, z = 3;
    y = y+ x*32;
    z = y+ x*x;
}

Then, compile that on Compiler Explorer.

Result is:

x:
        mov     r1, #1
        mov     r2, #2
        mov     r3, #3
        sub     sp, sp, #16
        str     r1, [sp, #4]
        str     r2, [sp, #8]
        str     r3, [sp, #12]
        ldr     r2, [sp, #4]
        ldr     r3, [sp, #8]
        add     r3, r3, r2, lsl #5
        str     r3, [sp, #8]
        ldr     r0, [sp, #4]
        ldr     r1, [sp, #4]
        ldr     r2, [sp, #8]
        mla     r3, r1, r0, r2
        str     r3, [sp, #12]
        add     sp, sp, #16
        bx      lr

After that, get the assignment of variables from the values. It looks like [sp, #4] is x, [sp, #8] is y, and [sp, #12] is z.

Finally, using this relation, construct the result.

The answer is:

add r2, r2, r1, lsl #5
mla r3, r1, r1, r2