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ARM 组装难题

发布于 2024-09-04 03:44:07 字数 208 浏览 3 评论 0原文

首先，我不确定是否存在解决方案。我花了几个多小时试图想出一个，所以要小心。

问题：

r1 包含任意整数，标志未根据其值设置。如果 r1 为 0x80000000，则将 r0 设置为 1，否则设置为 0，仅使用两条指令。

用 3 条指令很容易做到这一点（有很多方法），但是用 2 条指令做到这一点似乎非常困难，而且很可能是不可能的。

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云之铃。 2024-09-11 03:44:08

这是一个部分解决方案，它在 r0 的最高位给出了正确的答案，因此它可用作移位操作数 (r0 lsr #31)。

; r0 = r1 & -r1
rsb r0, r1, #0
and r0, r0, r1

这是有效的，因为 0 和 0x80000000 是唯一在求反时保留其符号位的数字。我相当确定精确的解决方案是不可能的。

编辑：不，这并非不可能。请参阅马丁的回答。

Here's a partial solution that gives the correct answer in the top bit of r0, so it is available as a shifter operand (r0 lsr #31).

; r0 = r1 & -r1
rsb r0, r1, #0
and r0, r0, r1

This works because 0 and 0x80000000 are the only numbers that retain their sign bits when negated. I'm fairly sure an exact solution is impossible.

EDIT: no, it's not impossible. See Martin's answer.

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情未る 2024-09-11 03:44:08

adds  r0, r1, #0x80000000 ; if r1 is 0x80000000, r0 will now hold 0
movne r0, #1              ; otherwise, set r0 to 1

这相当于：

unsigned int r0, r1;
r0 = r1 + 0x80000000; // 32 bits, so top bit will be lost on overflow
if (r0 != 0)
{
    r0 = 1;
}

adds  r0, r1, #0x80000000 ; if r1 is 0x80000000, r0 will now hold 0
movne r0, #1              ; otherwise, set r0 to 1

This is equivalent to:

unsigned int r0, r1;
r0 = r1 + 0x80000000; // 32 bits, so top bit will be lost on overflow
if (r0 != 0)
{
    r0 = 1;
}

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著墨染雨君画夕 2024-09-11 03:44:08

类似于：

mov r0,r1,lsr #31

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风柔一江水 2024-09-11 03:44:08

如果想使用“快速”指令，这是一个难题。我不太能想出一个解决方案，但可以提供更多“想法”：

; If goal were to have value of zero if $80000000 and something else otherwise:
  adds r0,r1,r1 ; Overflow only if $80000000
  movvc r0,#whatever

; If goal were to have value of $80000000 if $80000000 and zero otherwise
  subs r0,r1,#0  ; Overflow only if $80000000
  movvc r0,#0 ; Or whatever

; If the goal were to have value of $7FFFFFFF if $80000000 and zero otherwise
  adds r0,r1,r1,asr #31 ; Overflow only if $80000000
  movvc r0,#0

; If carry were known to be set beforehand
  addcs r0,r1,r1 ; Overflow only if $80000000 (value is 1)
  movvc r0,#0

; If register r2 were known to hold #1
  adds r0,r1,r1,asr #31
  ; If $80000000, MSB and carry set
  sbc r0,r2,r0,lsr #31

这些都不是完美的解决方案，但它们很有趣。

Tough puzzle if one wants to use "fast" instructions. I can't quite come up with a solution, but can offer a couple more 'notions':

; If goal were to have value of zero if $80000000 and something else otherwise:
  adds r0,r1,r1 ; Overflow only if $80000000
  movvc r0,#whatever

; If goal were to have value of $80000000 if $80000000 and zero otherwise
  subs r0,r1,#0  ; Overflow only if $80000000
  movvc r0,#0 ; Or whatever

; If the goal were to have value of $7FFFFFFF if $80000000 and zero otherwise
  adds r0,r1,r1,asr #31 ; Overflow only if $80000000
  movvc r0,#0

; If carry were known to be set beforehand
  addcs r0,r1,r1 ; Overflow only if $80000000 (value is 1)
  movvc r0,#0

; If register r2 were known to hold #1
  adds r0,r1,r1,asr #31
  ; If $80000000, MSB and carry set
  sbc r0,r2,r0,lsr #31

None of those is a perfect solution, but they're interesting.

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