第一个汇编程序错误“对mov的引用太多”

时间:2011-10-11 23:40:56

标签: assembly bit-manipulation

在下面的汇编代码中,我试图使用按位移位等实现乘法方法但是我一遍又一遍地得到两个错误,“'mov'的内存引用太多”以及“cmp的模糊操作数大小” /和/ MOV / SHL / SHR'”。有什么想法吗?

谢谢你们。根据要求,相关错误如下。

    .intel_syntax
    .data

    .globl x
x:  .long 0

    .globl y
y:  .long 0
    .text

    .globl multiply
multiply:
    push    ebp #These two statements are necessary
    mov ebp,esp

    mov eax,0
    mov ebx,x
    mov edx,y

LOOP:
    cmp ebx,0
    je  DONE

    mov ecx,ebx
    and ecx,1
    cmp ecx,1
    jne LOOPC   #jump to continued loop if low_bit(x) not 1

    add eax,edx

LOOPC:
    shr ebx,1
    shl edx,1
    jmp LOOP

DONE:
    pop ebp #Necessary statement
    ret     #return

错误讯息:

multiply.s: Assembler messages:
multiply.s:0: Warning: end of file in comment; newline inserted
multiply.s:15: Error: too many memory references for `mov'
multiply.s:17: Error: ambiguous operand size for `mov'
multiply.s:18: Error: too many memory references for `mov'
multiply.s:19: Error: too many memory references for `mov'
multiply.s:22: Error: ambiguous operand size for `cmp'
multiply.s:25: Error: too many memory references for `mov'
multiply.s:26: Error: ambiguous operand size for `and'
multiply.s:27: Error: ambiguous operand size for `cmp'
multiply.s:30: Error: too many memory references for `add'
multiply.s:33: Error: ambiguous operand size for `shr'
multiply.s:34: Error: ambiguous operand size for `shl'

1 个答案:

答案 0 :(得分:7)

您在所有注册名称前面缺少%

所以它应该是:

mov %eax,0
mov %ebx,x
mov %edx,y

实际上,它将它们解析为内存位置的变量。 (不包含%,它们与xy相同,其中内存引用。)

编辑:

啊......看起来ughoavgfhw已经在评论中指出了这一点。