For the LEGv8 assembly instructions below, what is 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 X 7, respectively. LSL X9. XO. #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]

Response: B[g]= A[f]+ A[f + 1];

Explanation:

• The very first command "LSL, X9, X0, #3" was to multiply the factor "f" by "8" use the shift function, this really is essential to get 8-byte words from the arrays.  

• The next command was to enter the array element "A" with index "f," the index of A[f] would be in' X9' from the first instruction implementation.

• The third and fourth instructions are the same as first and second statements however the array will be "B" and the index will be "g"

• The fifth statement "LDUR X0,[ X9, #0]" will assign the array value in "A[f]" to the variable "f" located in X0.

• The sixth command "ADDI X11, X9, #8" should lead in the next array value being stored after the "X11" address "A[f]," which is the value of the item A[f+1].  

• The next command "LDUR X9,[ X11,#0]" would be to load the command to "X9."  

• The next "ADD X9, X9, X0" instruction results in the addition of X0 with X9 contents, i.e. A[f] and A[f+1] array contents, resulting in X9 register.  

• During the last request, the value stored in X9 is stored in X10, assigned to the address of B[ g] • Therefore, the overall operation performed by the instructions given is: B[g]= A[f]+ A[f + 1]

Therefore, the related C statement is B[g]= A[f]+ A[f + 1].

answer: a, c

divide any of the y values by x values and you will always get 3 is the most simple way i can put it for you.

no, it just either gets automatically 5 points to give away, or u can choose how many points u want to give but, the answerer only gets half