我正在尝试将此递归Fibonacci代码转换为arm汇编语言。我是新手,并不确定如何做到这一点。我有一些我在下面玩过的代码片段。
Fib (n) {
if (n == 0 || n == 1) return 1;
else return Fib(n-2) + Fib(n-1);
}
到目前为止,这是我的尝试:
RO = 1
CMP RO #1
BGT P2
MOV R7 #1
B END
P2:
END LDR LR [SO,#0]
ADD SP SP, #8
MOV PC, LR
非常感谢帮助
答案 0 :(得分:1)
为了避免吃勺,我写了一个LEGv8程序,使用递归找到Fibonacci序列。 LEGv8 略微与ARMv8不同,但算法仍然存在。
请查看代码,并将命令/寄存器更改为ARMv8中的相应值。
我假设n(Fibonacci序列的范围)存储在寄存器X19中。
我还假设我们应该将Fibonacci序列存储在一个数组中,该数组的基地址存储在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