解释一行二进制补码数学计数空闲虚拟内存

Question

我有一个适用于 Windows Mobile 6.x 的 Visual Studio 2008 C++ 应用程序，我在其中计算给定进程可用的可用虚拟内存量。 （我意识到它没有考虑碎片。）我的代码基本上如下所示：

MEMORY_BASIC_INFORMATION mbi = { 0 };

/// total free memory available to the process
DWORD free = 0;

/// base memory address for the given process index (2-33). 
DWORD slot_base_addr = process_index * 0x02000000;

/// look at each memory region for the process. 
for( DWORD offset = 0x10000; 
     offset < 0x02000000; 
     offset += mbi.RegionSize )
{
    ::VirtualQuery( ( void* )( slot_base_addr + offset ), 
                    &mbi, 
                    sizeof( MEMORY_BASIC_INFORMATION ) );

    if( mbi.State == MEM_FREE )
    {
        free += ( mbi.RegionSize - ( ( ~( DWORD )mbi.BaseAddress + 1 ) & 0xffff ) ) & 0xffff0000;
    }
}
NKDbgPrintfW( L"%d bytes free\r\n", free );

我可以通过其他 API 确认这似乎工作正常。我的问题是这一行在做什么：

free += ( mbi.RegionSize - ( ( ~( DWORD )mbi.BaseAddress + 1 ) & 0xffff ) ) & 0xffff0000;

为什么这不仅仅是：

free += mbi.RegionSize;

我在 Usenet 由 MSFT 员工 Ross Jordan 发布。

谢谢， PaulH

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编辑：

例如。对于进程槽 2，这是每个空闲内存块的列表，其中空闲内存量由 Ross Jordan (RS) 算法和 RegionSize (RS) 给出。

Slot: 2. Range: 0x04000000 - 0x06000000
    RS:    16,384 bytes RJ:         0 bytes diff: 16384
    RS:     4,096 bytes RJ:         0 bytes diff: 4096
    RS:     4,096 bytes RJ:         0 bytes diff: 4096
    RS:     4,096 bytes RJ:         0 bytes diff: 4096
    RS:     4,096 bytes RJ:         0 bytes diff: 4096
    RS:     4,096 bytes RJ:         0 bytes diff: 4096
    RS:    36,864 bytes RJ:         0 bytes diff: 36864
    RS:    65,536 bytes RJ:    65,536 bytes diff: 0
    RS:    53,248 bytes RJ:         0 bytes diff: 53248
    RS:     4,096 bytes RJ:         0 bytes diff: 4096
    RS:     4,096 bytes RJ:         0 bytes diff: 4096
    RS:     4,096 bytes RJ:         0 bytes diff: 4096
    RS:     4,096 bytes RJ:         0 bytes diff: 4096
    RS:     4,096 bytes RJ:         0 bytes diff: 4096
    RS: 7,671,808 bytes RJ: 7,667,712 bytes diff: 4096
    RS: 1,921,024 bytes RJ: 1,900,544 bytes diff: 20480
    RS: 7,491,584 bytes RJ: 7,471,104 bytes diff: 20480
    RS: 3,252,224 bytes RJ: 3,211,264 bytes diff: 40960
    RS:   262,144 bytes RJ:   262,144 bytes diff: 0

RS: Total VM Free: 20,811,776 bytes.
RJ: Total VM Free: 20,578,304 bytes.

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编辑2：

汉斯引导我找到了答案。这只是一种奇特的方法，但假设分配大小为 64KB。

SYSTEM_INFO si = { 0 };
::GetSystemInfo( &si );

free += mbi.RegionSize - mbi.RegionSize % si.dwAllocationGranularity;

Answer 1

VirtualAlloc 的分配粒度通常为 64KB。如果 AllocationBase 不是 64KB 的倍数，他会尝试做一些有意义的事情。我认为这根本没有意义，他的位掩码仍然假设粒度为 64KB，并且他不使用 SYSTEM_INFO.dwAllocationGranularity。其中有这样的评论：

该值被硬编码为 64 KB
过去，但其他硬件
架构可能需要不同的
值。

在极少数情况下（不是 64KB），此代码将生成垃圾值。只需按 RegionSize 进行操作即可。