下面的代码(在C ++中)是我尝试转换为C#
的代码DWORD Func_X_4(DWORD arg1, DWORD arg2, DWORD arg3)
{
LARGE_INTEGER result = {1, 0};
LARGE_INTEGER temp1 = {0};
LARGE_INTEGER temp2 = {0};
LARGE_INTEGER temp3 = {0};
LARGE_INTEGER temp4 = {0};
for(int x = 0; x < 32; ++x)
{
if(arg2 & 1)
{
temp1.LowPart = arg3;
temp1.HighPart = 0;
temp2.QuadPart = temp1.QuadPart * result.QuadPart;
temp3.LowPart = arg1;
temp3.HighPart = 0;
temp4.QuadPart = temp2.QuadPart % temp3.QuadPart;
result.QuadPart = temp4.QuadPart;
}
arg2 >>= 1;
temp1.LowPart = arg3;
temp1.HighPart = 0;
temp1.QuadPart *= temp1.QuadPart;
temp2.LowPart = arg1;
temp2.HighPart = 0;
temp3.QuadPart = temp1.QuadPart % temp2.QuadPart;
arg3 = temp3.LowPart;
if(!arg2)
break;
}
return result.LowPart;
}
在这里,我尝试翻译代码,但它太乱了,我之前从未使用过大整数。
结构:
public struct LARGE_INTEGER
{
UInt32 LowPart;
Int32 HighPart;
Int32 QuadPart;
}
翻译功能:
public Int32 Func_X_4(Int32 arg1, Int32 arg2, Int32 arg3)
{
LARGE_INTEGER result = {1, 0}; //this and the four below,are they correct?
LARGE_INTEGER temp1 = {0, 0};
LARGE_INTEGER temp2 = {0, 0};
LARGE_INTEGER temp3 = {0, 0};
LARGE_INTEGER temp4 = {0, 0};
for(int x = 0; x < 32; ++x)
{
if(arg2 & 1==0) //correct?
{
temp1.LowPart = arg3;
temp1.HighPart = 0;
temp2.QuadPart = temp1.QuadPart * result.QuadPart;
temp3.LowPart = arg1;
temp3.HighPart = 0;
temp4.QuadPart = temp2.QuadPart % temp3.QuadPart;
result.QuadPart = temp4.QuadPart;
}
arg2 >>= 1;
temp1.LowPart = arg3;
temp1.HighPart = 0;
temp1.QuadPart *= temp1.QuadPart;
temp2.LowPart = arg1;
temp2.HighPart = 0;
temp3.QuadPart = temp1.QuadPart % temp2.QuadPart;
arg3 = temp3.LowPart;
if(arg2 != 0) //correct?
break;
}
return result.LowPart;
}
问题: 第一个问题是我没有在C#中找到一个LARGE_INTEGER类型变量,所以我创建了一个结构,我想知道是否确实存在。 至于第二个问题,功能不对,没有用。
对这个特定问题的任何帮助都将非常感谢! 提前谢谢。
答案 0 :(得分:11)
LARGE_INTEGER结构的直接翻译是:
[StructLayout(LayoutKind.Explicit, Size=8)]
struct LARGE_INTEGER
{
[FieldOffset(0)]public Int64 QuadPart;
[FieldOffset(0)]public UInt32 LowPart;
[FieldOffset(4)]public Int32 HighPart;
}
它就像C中的联合,其中QuadPart
是一个8字节的值,LowPart
占据前4个字节而HighPart
占据高位4个字节。
答案 1 :(得分:5)
其Int64。基于http://msdn.microsoft.com/en-us/library/aa383713.aspx,它是一个64位有符号整数。
答案 2 :(得分:1)
在前一个答案的基础上,我发现扩展LARGE_INTEGER的定义以包含高和低部分的int和uint版本是有用的:
[StructLayout(LayoutKind.Explicit, Size=8)]
struct LARGE_INTEGER
{
[FieldOffset(0)]public long QuadPart;
[FieldOffset(0)]public uint LowPart;
[FieldOffset(4)]public int HighPart;
[FieldOffset(0)]public int LowPartAsInt;
[FieldOffset(0)]public uint LowPartAsUInt;
[FieldOffset(4)]public int HighPartAsInt;
[FieldOffset(4)]public uint HighPartAsUInt;
}