为什么这个添加的byte*和uint无法进入更高的dword？

Question

现在提交Microsoft Connect ; 如果您觉得需要修理请立即投票.我也简化了测试用例:

byte* data = (byte*) 0x76543210;
uint offset = 0x80000000;
byte* wrong = data + offset;
byte* correct = data + (uint) 0x80000000;

// "wrong" is now 0xFFFFFFFFF6543210 (!)
// "correct" is 0xF6543210

看看IL,就我所知,C#编译器做的一切都正确,错误在于JITter.

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原始问题:这里发生了什么？

byte* data = (byte*)Marshal.AllocHGlobal(0x100);

uint uioffset = 0xFFFF0000;
byte* uiptr1 = data + uioffset;
byte* uiptr2 = data + (uint)0xFFFF0000;

ulong uloffset = 0xFFFF0000;
byte* ulptr1 = data + uloffset;
byte* ulptr2 = data + (ulong)0xFFFF0000;

Action<string, ulong> dumpValue =
    (name, value) => Console.WriteLine("{0,8}: {1:x16}", name, value);

dumpValue("data",     (ulong)data);
dumpValue("uiptr1",   (ulong)uiptr1);
dumpValue("uiptr2",   (ulong)uiptr2);
dumpValue("ulptr1",   (ulong)ulptr1);
dumpValue("ulptr2",   (ulong)ulptr2);

此测试需要针对x64平台的64位操作系统.

输出:

  data: 000000001c00a720    (original pointer)
uiptr1: 000000001bffa720    (pointer with a failed carry into the higher dword)
uiptr2: 000000011bffa720    (pointer with a correct carry into the higher dword)
ulptr1: 000000011bffa720    (pointer with a correct carry into the higher dword)
ulptr2: 000000011bffa720    (pointer with a correct carry into the higher dword)
               ^
               look here

那么这是一个错误还是我搞砸了什么？

Answer 1

我认为您遇到了这个 C# 编译器错误：https://connect.microsoft.com/VisualStudio/feedback/details/675205/c-compiler-performs-sign-extension-during-unsigned-pointer-arithmetic

这是由于以下问题而提交的：64-bitpointer Algorithm in C#, Check for算术溢出更改行为