VS:_BitScanReverse64 内在的意外优化行为

VS: unexpected optimization behavior with _BitScanReverse64 intrinsic(VS:_BitScanReverse64 内在的意外优化行为)

问题描述

以下代码在调试模式下工作正常，因为 _BitScanReverse64 已定义如果没有设置位，则返回 0.引用 MSDN:(返回值是)如果设置了索引，则非零，如果未找到设置位，则为 0."

The following code works fine in debug mode, since _BitScanReverse64 is defined to return 0 if no Bit is set. Citing MSDN: (The return value is) "Nonzero if Index was set, or 0 if no set bits were found."

如果我在发布模式下编译这段代码它仍然有效，但如果我启用编译器优化，例如 O1 或 O2 索引不为零并且 assert() 失败.

If I compile this code in release mode it still works, but if I enable compiler optimizations, such as O1 or O2 the index is not zero and the assert() fails.

#include <iostream>
#include <cassert>

using namespace std;

int main()
{
  unsigned long index = 0;
  _BitScanReverse64(&index, 0x0ull);

  cout << index << endl;

  assert(index == 0);

  return 0;
}

这是预期的行为吗?我使用的是 Visual Studio Community 2015，版本 14.0.25431.01 更新 3.(我留下了 cout，以便在优化过程中不会删除变量索引).还有没有有效的解决方法，还是我不应该直接使用这个编译器内在函数?

Is this the intended behaviour ? I am using Visual Studio Community 2015, Version 14.0.25431.01 Update 3. (I left cout in, so that the variable index is not deleted during optimization). Also is there an efficient workaround or should I just not use this compiler intrinsic directly?

推荐答案

AFAICT，intrinsic 在输入为零时在 index 中留下垃圾， 弱于行为asm 指令.这就是为什么它有一个单独的布尔返回值和整数输出操作数.

AFAICT, the intrinsic leaves garbage in index when the input is zero, weaker than the behaviour of the asm instruction. This is why it has a separate boolean return value and integer output operand.

尽管 index 参数是通过引用获取的，但编译器将其视为仅输出.

Despite the index arg being taken by reference, the compiler treats it as output-only.

unsigned char _BitScanReverse64 (unsigned __int32* index, unsigned __int64 mask)
英特尔针对相同内在函数的内在函数指南文档似乎比您链接的 Microsoft 文档 更清晰，并阐明了 MS文档试图说.但是仔细阅读，它们似乎都在说同样的事情，并且描述了围绕 bsr 指令的薄包装.

unsigned char _BitScanReverse64 (unsigned __int32* index, unsigned __int64 mask)
Intel's intrinsics guide documentation for the same intrinsic seems clearer than the Microsoft docs you linked, and sheds some light on what the MS docs are trying to say. But on careful reading, they do both seem to say the same thing, and describe a thin wrapper around the bsr instruction.

英特尔记录了BSR 指令当输入为 0 时产生未定义的值"，但在这种情况下设置 ZF. 但 AMD 将其记录为保持目标不变:

Intel documents the BSR instruction as producing an "undefined value" when the input is 0, but setting the ZF in that case. But AMD documents it as leaving the destination unchanged:

AMD 的 BSF 条目 AMD64 架构程序员手册第 3 卷:通用和系统说明

... 如果第二个操作数包含 0，则指令设置 ZF为 1 并且不改变目标寄存器的内容....

... If the second operand contains 0, the instruction sets ZF to 1 and does not change the contents of the destination register. ...

在当前的 Intel 硬件上，实际行为与 AMD 的文档相符:当 src 操作数为 0 时，它使目标寄存器保持不变.也许这就是为什么 MS 将其描述为仅在输入为时设置 Index非零(并且内部函数的返回值非零).

On current Intel hardware, the actual behaviour matches AMD's documentation: it leaves the destination register unmodified when the src operand is 0. Perhaps this is why MS describes it as only setting Index when the input is non-zero (and the intrinsic's return value is non-zero).

在英特尔(但可能不是 AMD)上，就这样因为甚至没有将 64 位寄存器截断为 32 位.例如mov rax,-1 ;bsf eax, ecx(ECX 为零)使 RAX=-1(64 位)，而不是您从 xor eax, 0<得到的 0x00000000ffffffff/代码>.但是对于非零 ECX，bsf eax, ecx 具有将零扩展到 RAX 的通常效果，例如留下 RAX=3.

if(_BitScanReverse64(&idx, input)) {
    // idx is valid.
    // (MS docs say "Index was set")
} else {
    // input was zero, idx holds garbage.
    // (MS docs don't say Index was even set)
    idx = -1;     // might make sense, one lower than the result for bsr(1)
}