爲了避免填鴨式的緣故,我寫道,發現使用遞歸斐波那契序列的LEGv8程序。 LEGv8是略微不同於ARMv8,但算法仍然存在。
請檢查代碼,並將命令/寄存器更改爲其在ARMv8中的相應值。
我假設n(Fibonacci序列的範圍)存儲在寄存器X19中。
我還假設我們應該將Fibonacci序列存儲在一個數組中,該數組的基地址存儲在X20中。
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
非常感謝。當我將它轉換爲ARM時,我會盡快與您聯繫。 –