如何在 C# 中使用 & 运算符？代码的翻译是否正确？
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How to use the & operator in C#? Is the the translation of the code correct?
提问by Ivan Prodanov
the line "if(arg2 & 1)" in C++(arg2 is DWORD) is equal to "if(arg2 & 1==0)" in C#(arg2 is Uint32),right?
C++ 中的“if(arg2 & 1)”行（arg2 是 DWORD）等于 C# 中的“if(arg2 & 1==0)”（arg2 是 Uint32），对吗？
I am trying to translate a function from C++ to C#,but I get an error:
我正在尝试将函数从 C++ 转换为 C#，但出现错误：
Operator '&' cannot be applied to operands of type 'uint' and 'bool'
I'd be also thankful if you could see further in the whole function for any other mistakes.
如果您能在整个函数中进一步了解任何其他错误，我也将不胜感激。
C++
C++
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;
}
Converted to C#
转换为 C#
LARGE_INTEGER structure:
LARGE_INTEGER 结构：
[StructLayout(LayoutKind.Explicit, Size = 8)]
public struct LARGE_INTEGER
{
[FieldOffset(0)]
public Int64 QuadPart;
[FieldOffset(0)]
public UInt32 LowPart;
[FieldOffset(4)]
public Int32 HighPart;
}
Function:
功能：
public static UInt32 X4(UInt32 arg1, UInt32 arg2, UInt32 arg3)
{
LARGE_INTEGER result = new LARGE_INTEGER();
result.LowPart = 1;
result.HighPart = 0;
LARGE_INTEGER temp1 = new LARGE_INTEGER();
LARGE_INTEGER temp2 = new LARGE_INTEGER();
LARGE_INTEGER temp3 = new LARGE_INTEGER();
LARGE_INTEGER temp4 = new LARGE_INTEGER();
for (int x = 0; x < 32; ++x)
{
if (arg1 & 1 ==0)
{
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)
break;
}
return result.LowPart;
}
This is what I'm not sure yet:
这是我还不确定的：
- Whether a DWORD in C++ is UInt32 or Int32 in C#?
- if(integer & integer) means if(integer and integer ==0)? //this is where the error i described above is placed.
- if(!integer) means if(integer != 0)?
- Why operator & cannot be used logically in C# ,meaning it requires a boolean?
- "LARGE_INTEGER result = {1, 0}" means result.lowpart is 1 and result.highpart is 0 or result.Quadpart = 1?
- C++ 中的 DWORD 是 UInt32 还是 C# 中的 Int32？
- if(integer & integer) 表示 if(integer and integer ==0)？//这是我上面描述的错误所在的位置。
- if(!integer) 表示 if(integer != 0)？
- 为什么运算符 & 不能在 C# 中逻辑使用，这意味着它需要一个布尔值？
- "LARGE_INTEGER result = {1, 0}" 表示 result.lowpart 为 1，result.highpart 为 0 或 result.Quadpart = 1？
Thanks in advance!
提前致谢！
采纳答案by ybo
Where you write :
你在哪里写：
if (arg1 & arg2==0)
如果 (arg1 & arg2==0)
The compiler understands :
编译器理解：
if (arg1 & (arg2==0))
You should write :
你应该写：
if ((arg1 & arg2) == 0)
This is the way the C++ statement should be translated to C# :
这是将 C++ 语句转换为 C# 的方式：
if (arg2 & 1) // C++ (arg2 is DWORD)
if ((arg2 & 1) != 0) // C# (arg2 is Uint32)
Or, in a more generic way:
或者，以更通用的方式：
if (Flags & FlagToCheck) // C++
if ((Flags & FlagToCheck) != 0) // C#
In C/C++, 0 is false, everything else is true.
在 C/C++ 中，0 为假，其他均为真。
- You should check the definition of DWORD, it should be (unsigned int), which is UInt32 in C#
- if (integer & integer), in C/C++ means "if the result of the bitwise and between the two integers is not 0" (0 is false, everything else is true).
- if (!integer) means if (integer == 0) (again, 0 is false, everything else is true)
- in C#, like in Java I think, booleans and numbers are two different things, you can only use booleans in "if" statements, there is not implicit conversion if you use an int : it won't compile.
- I'll leave this one to someone else, I'd need to test to be sure...
- 你应该检查DWORD的定义，它应该是(unsigned int)，也就是C#中的UInt32
- if (integer & integer)，在 C/C++ 中的意思是“如果按位和两个整数之间的结果不是 0”（0 为假，其他一切都为真）。
- if (!integer) 表示 if (integer == 0) （同样，0 是假的，其他一切都是真的）
- 在 C# 中，就像在 Java 中一样，我认为布尔值和数字是两种不同的东西，只能在“if”语句中使用布尔值，如果使用 int 则没有隐式转换：它不会编译。
- 我会把这个留给别人，我需要测试以确保......
回答by strager
DWORDisuint32_tin C++, thusUInt32in C#.if(a & b)converts toif((a & b) != 0).!=is evaluated before&thus the&expression needs parentheses around it.if(x)converts toif(x != 0)&is a 'bitwise and' in C#, like in C++.- Depends on your C++ structure.
DWORD是uint32_t在C ++，因此UInt32在C＃。if(a & b)转换为if((a & b) != 0).!=之前被评估，&因此&表达式需要在它周围加上括号。if(x)转换为if(x != 0)&在 C# 中是一个“按位与”，就像在 C++ 中一样。- 取决于您的 C++ 结构。
回答by Jim Mischel
5 - It means both. Because LowPart and HighPart are just "windows" into QuadPart's memory, when result.LowPart == 1 and Result.HighPart == 0, then result.QuadPart will be equal to 1.
5 - 两者都有。因为 LowPart 和 HighPart 只是 QuadPart 内存中的“窗口”，当 result.LowPart == 1 和 Result.HighPart == 0 时，result.QuadPart 将等于 1。
