ARM程序集使用递归和堆栈计算x ^ n

时间:2019-03-26 23:35:18

标签: recursion assembly arm stack

我有一项作业,要求我使用递归计算x ^ n。提供了功能power的C源代码,我必须将其翻译为汇编语言,但是我是汇编的新手,有点卡住了,所以不确定该怎么做。

到目前为止,这是我的代码:

        AREA powerXn, CODE, READONLY
        ENTRY

        MOV r0, #2                      ;Value of x.
        MOV r1, #3                      ;Value of n.
        ADR sp, stack                   ;Points sp to start of stack.
        BL main
loop    B loop

main    STMFD sp!, {r0, r1, r2, r3, lr} ;Pushes registers on to stack.
        BL power
        LDMFD sp!, {r0, r1, r2, r3, pc} ;Pops values from stack.

power   STMFD sp!, {r0, r1, lr}         ;Pushes registers on to stack.
        CMP r1, #0                      ;Checks the base case
        MOVEQ r2, #1                    ;Sets r2 to be 1 if base case is reached
        BNE test                        ;If base case not reached, jump to test.

        MUL r3, r0, r2              
        LDMFD sp!, {r0, r1, pc}

test    TST r1, #1                      ;Checks if r1 (n) is odd
        BNE odd                         ;If it is odd, jump to odd
        BEQ even                        ;Otherwise jump to even

odd     SUB r1, r1, #1                  ;Subtract 1 from r1 (n).
        BL power                        ;Recurse back to power.
        MUL r3, r0, r2              
        LDMFD sp!, {r0, r1, pc}

even    ;Not sure what to do here and not sure if what i have done so far is right.


        AREA powerXn, DATA, READWRITE
        ALIGN
        space 0x200
stack   DCD 0x00
        END

我需要翻译的C语言源代码是:

int power(int x, unsigned int n){
 int y;

 if (n == 0)
      return 1;

 if (n & 1) // if n is odd
      return x * power(x, n - 1);
 else { // if n is even  
      y = power(x, n >> 1); // n >> 1 means n/2
      return y * y;
 }
}

1 个答案:

答案 0 :(得分:1)

您似乎对C更加熟悉,所以我建议转到https://godbolt.org/并尝试使用C到汇编器功能。您所要做的就是选择一个正确的编译器(ARM GCC有一些选项)并浏览汇编代码。当然,我强烈反对简单地复制代码而不先尝试理解它,但是该网站应该至少为您提供有关操作的线索。我目前正在与FASM学习x86汇编程序,该网站是我进行研究期间的首选之地。