Question

Convert the following C statement to the corresponding ARMv7 assembly code: Assume that the variables f, g, h, i, and j are assigned to registers X0,X1,X2,X3, and X4, respectively. Assume that the base address of the arrays A and B are in registers X6 and X7, respectively.

B[8] = A[i−j];

Convert the following LEGv8 assembly instructions to ARMv7 as well as find the corresponding C statement. Assume that the variables f, g, h, i, and j are assigned to registers X0,X1,X2,X3, and X4, respectively. Assume that the base address of the arrays A and B are in registers X6 and X7, respectively.

LSL X9, X0, #3 // X9 = f*8

ADD X9, X6, X9 // X9 = &A[f]

LSL X10, X1, #3 // X10 = g*8

ADD X10, X7, X10 // X10 = &B[g]

LDUR X0, [X9, #0]. // f = A[f]

ADDI X11, X9, #8

LDUR X9, [X11, #0]

ADD X9, X9, X0

STUR X9, [X10, #0]

Answer 1

Solution

B[g] = A[f] + A[f + 1]

Explanation

we can clearly see that A[f] is added A[f+1] and

result is stored on

B[g]

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all the best