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nasm idiv a negative value (1 answer)

Closed 6 months ago.

;print out division message
mov rcx, 0                       ;zero out register
mov rax, [input]
mov rcx, [input2]
idiv rcx                        ;divide rax by rcx
mov rdi, rax                    ;for printing purposes
call print_int

我看不出,为什么这种n断,用64台借机将一米带“空洞点例外”,而且这些数值不是浮动点。

我知道,在进行分裂之后,引人应当坐在黑手里,而其余的人应当相信,但从现在起,我只是想把我的手带上来。

Answer 1

你的职能对我来说是很困难的。 www.un.org/Depts/DGACM/index_spanish.htm

_mydiv:
  xor  %rdx, %rdx ; clear high bits of dividend
  mov  %rdi, %rax ; copy dividend argument into rax
  idiv %rsi       ; divide by divisor argument
  ret             ; return (quotient is in rax)

Translated into NASM syntax and to the windows ABI, I think that would be something like:

_mydiv:
  mov  r8, rdx    ; copy divisor argument to scratch register
  xor  rdx, rdx   ; clear high bits of dividend
  mov  rax, rcx   ; copy dividend argument into rax
  idiv r8         ; divide by divisor in scratch register
  ret             ; return (quotient is in rax)

您的参数是否一致,如何混淆一些东西?

Edit: looking at your code, it occurs to me that it might not be written as a proper function at all. The important steps are:

Put dividend in RDX:RAX - for you that probably means clearing out RDX and putting the input dividend in RAX.
Put divisor in some other register - you chose RCX, that should be fine.
Divide - idiv rcx.
Result will be in RAX.

You should pay particular attention to step 1 - make sure that RDX:RAX has sane contents! Why you re getting a floating point exception I can t guess from the code you ve shown.

Answer 2

您实际上将RDX:RAX的128倍数字由RCX区分。因此,如果RDX未启动,结果可能大于64倍,导致出现超支例外。在该司之前添加一个CQO,以将RAS输入RDX。

我可以解释一下浮动点的比值,但也许有人决定重新使用通用数学差错的中断矢量,有些地方倒下了线?

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