Question

4. Convert the below C code snippet to LEGV8 assembly code. Base address of x is stored in register X19. Assume variables a, b, andc are stored in registers X20, X21, and X22 respectively. Assume all values are 64-bits. Do not use divide and multiply instructions in your code. Comment your assembly code. (30 Points) x[e] a x[1]; q [e]x x[a/2]b; x[2]; + x[1] x[2] x[c] C >> 4: x[1] +

Answer 1

ld r1, 8[X19] /* value of x[1] is stored in r1 register (as each integer takes 8 Bytes, X[1] = *(X+8)) */
sub r0, X20, r1 /* r0 := a - x[1] */
st r0, [X19] /* store r0 value into place whose address is X19 i.e., x[0] */
add X22, r0, X21 /* c := x[0] + b */
lsl r2, c, 4 /* r2 := c >> 4 */
st r2, 8[X19] /* store r2 value into place whose address is X19+8 i.e., x[1] */
mul r4, X20, 4 /* r4 := a * 4 */
ld r5, r4[X19] /* value from the address X19+r4 = x[r4/8] = x[a/2] is loaded into r5 */
sum r3, r5, X21 /* r3 := r5 + b */
st r3, 16[X19] /* store r3 value into place whose address is X19+16 i.e., x[2] */
sum r6, r1, r3 /* r6 := x[1] + x[2] */
mul r7, X22, 8 /* r7 := c * 8 */
st r6, r7[X19] /* r5 is stored into place whose address is X19+r7 = x[r7/8] = x[c] */

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