|=（单管道相等）和 &=（单 & 符号相等）是什么意思
在下面几行中：
//Folder.Attributes = FileAttributes.Directory | FileAttributes.Hidden | FileAttributes.System | FileAttributes.ReadOnly;
Folder.Attributes |= FileAttributes.Directory | FileAttributes.Hidden | FileAttributes.System | FileAttributes.ReadOnly;
Folder.Attributes |= ~FileAttributes.System;
Folder.Attributes &= ~FileAttributes.System;
|=（单管道相等）和 &=（单与号相等）在 C# 中意味着什么？
我想删除系统属性并保留其他属性......
In below lines:
//Folder.Attributes = FileAttributes.Directory | FileAttributes.Hidden | FileAttributes.System | FileAttributes.ReadOnly;
Folder.Attributes |= FileAttributes.Directory | FileAttributes.Hidden | FileAttributes.System | FileAttributes.ReadOnly;
Folder.Attributes |= ~FileAttributes.System;
Folder.Attributes &= ~FileAttributes.System;
What does |= (single pipe equal) and &= (single ampersand equal) mean in C#?
I want to remove system attribute with keeping the others...
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它们是 复合赋值运算符，翻译（非常松散）
为
和 与
&相同。在一些关于隐式转换的情况下有更多细节，并且目标变量仅被评估一次，但这基本上就是它的要点。就非复合运算符而言，
&是按位“AND” 和|是按位“OR”。编辑：在这种情况下，您需要
Folder.Attributes &= ~FileAttributes.System。要理解原因：~FileAttributes.System表示“除System之外的所有属性”（~是按位非)&表示“结果是操作数两侧出现的所有属性”，因此它基本上充当掩码 - 仅保留出现在 ( “除了系统之外的一切”）。一般来说：
|=只会向目标添加位&=只会删除位目标They're compound assignment operators, translating (very loosely)
into
and the same for
&. There's a bit more detail in a few cases regarding an implicit cast, and the target variable is only evaluated once, but that's basically the gist of it.In terms of the non-compound operators,
&is a bitwise "AND" and|is a bitwise "OR".EDIT: In this case you want
Folder.Attributes &= ~FileAttributes.System. To understand why:~FileAttributes.Systemmeans "all attributes exceptSystem" (~is a bitwise-NOT)&means "the result is all the attributes which occur on both sides of the operand"So it's basically acting as a mask - only retain those attributes which appear in ("everything except System"). In general:
|=will only ever add bits to the target&=will only ever remove bits from the target|是 按位或&是 按位与a |= b相当于a = a | b除了a仅计算一次a &= b相当于a = a & b除外，a仅计算一次。为了删除系统位而不更改其他位，请使用
~按位求反。因此，您可以将除系统位之外的所有位设置为 1。and-使用掩码将系统设置为 0 并保持所有其他位不变，因为0 & x = 0 和 1 & x = x对于任何x|is bitwise or&is bitwise anda |= bis equivalent toa = a | bexcept thatais evaluated only oncea &= bis equivalent toa = a & bexcept thatais evaluated only onceIn order to remove the System bit without changing other bits, use
~is bitwise negation. You will thus set all bits to 1 except the System bit.and-ing it with the mask will set System to 0 and leave all other bits intact because0 & x = 0and1 & x = xfor anyx您可以这样做：
You can do this like so: