Question

For the following C statement, write the corresponding LEGv8 assembly code.

long long int func( long long int n ) {

if ( n <= 1 ) {

return n;

}

return func( n – 1 ) + func( n – 2 );

}

Answer 1

I assumed that n is stored in register X19.

I assumed that store the Fibonacci sequence in an array, which has its base address stored in X20.

LEGv8 assembly code:

MOV X17, XZR // keep (previous) 0 in X17 for further use

ADDI X18, XZR, #1 // keep (Current) 1 in X18 for further use

ADDI X9, XZR, #0 // Assuming i = 0 is in register X9

fibo:

SUBI SP, SP, #24 // Adjust stack pointer for 3 items

STUR LR, [SP, #16] // save the return address

STUR X17, [SP, #8] //save content of X17 on the stack

STUR X18, [SP, #0] //save content of X18 on the stack

SUBS X10, X9, X19 // test for i==n CBNZ X10, L1 // If i not equal to n, go to L1

MOV X6, XZR // keep 0 on X6

ADDI X5, XZR, #1 // keep 1 on X5

ADDI X2, X9, #1 //X9 increased by 1 for further use

STUR X6, [X20,#0] //store 0 in the array

STUR X5, [X20, #8] //store 1 in the array

ADDI SP, SP, #24 // pop 3 items from the stack

BR LR // return to the caller

L1:

ADD X16, X17, X18 // Next_Num = previous + Current

MOV X17, X18 // Previous = Current

MOV X18, X16 // Current= Next_Num

ADDI X9, X9, #1 // i++ BL fibo // call fibo

LDUR X18, [SP, #0] // return from BL; restore previous

LDUR X17, [SP, #8] // restore current

LDUR LR, [SP, #16] // restore the return address

ADDI SP, SP, #24 // adjust stack pointer to pop 3 items

ADD X7, X18, X17 // keep (previous + current) value on register X7 LSL X2, X2, #3 // Multiplying by 8 for offset

ADD X12, X20, X2 // address of the array increase by 8

STUR X7, [X12, #0] // store (previous + current) value on the array

SUBI X2, X2, #1 // X9 decreased by 1

BR LR // return

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