C# 枚举标志比较
给定以下标志，
[Flags]
public enum Operations
{
add = 1,
subtract = 2,
multiply = 4,
divide = 8,
eval = 16,
}
我如何实现 IF 条件来执行每个操作？在我的尝试中，第一个条件对于add, eval成立，这是正确的。然而，第一个条件对于 subtract, eval 也成立，这是不正确的。
public double Evaluate(double input)
{
if ((operation & (Operations.add & Operations.eval)) == (Operations.add & Operations.eval))
currentResult += input;
else if ((operation & (Operations.subtract & Operations.eval)) == (Operations.subtract & Operations.eval))
currentResult -= input;
else
currentResult = input;
operation = null;
return currentResult;
}
我看不出问题是什么。
Given the following flags,
[Flags]
public enum Operations
{
add = 1,
subtract = 2,
multiply = 4,
divide = 8,
eval = 16,
}
How could I implement an IF condition to perform each operation? In my attempt, the first condition is true for add, eval, which is correct. However the first condition is also true for subtract, eval, which is incorrect.
public double Evaluate(double input)
{
if ((operation & (Operations.add & Operations.eval)) == (Operations.add & Operations.eval))
currentResult += input;
else if ((operation & (Operations.subtract & Operations.eval)) == (Operations.subtract & Operations.eval))
currentResult -= input;
else
currentResult = input;
operation = null;
return currentResult;
}
I cannot see what the problem is.
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评论（5）
将内部
&更改为|：这相当于：
这可能更具可读性。您可能还想考虑这样的扩展：
然后您可以这样做：
这可能更具可读性。最后，您可以创建这个扩展以获得更多乐趣：
然后您的表达式可以变成：
Change your inner
&to|:This is equivalent to:
which might be more readable. You might also want to consider an Extension like this:
then you can do this:
which might be even more readable. Finally you could create this extension for even more fun:
Then your expression could turn into:
哇，我无法相信所有错误的答案。
如果您使用标志，那么理解按位数学非常重要。在您的情况下，您有以下内容（对于第一个条件）：
因此，假设我们将减去（2）作为
操作，因为之前的结果是
00000任何AND' d 将为零。因此，由于0 == 0，您的条件将始终评估为true。如果我们只是将其切换为 OR，那么我们将得到以下结果：
现在，假设我们将
Add (1)作为操作所以，<代码>1 & 17 => 1 因此你的最终条件是
(1 & ( 1 | 16 ) ) == ( 1 | 16 )=> <代码>1 & 17 == 17 =>;1 == 17=>false（仍然是 false！）所以你真正想要的是：
这变成
((1 | 1 | 16 ) & ( 1 | 16 )) == ( 1 | 16 )=>( 17 & 17 ) == 17=>;17 == 17==true这显然不可读，因此您应该选择将其提取到方法中（按照建议）。但了解为什么您的情况不正确仍然很重要。
Wow, I can't believe all of the wrong answers..
It's important to understand bitwise math if you're working with flags. In your case, you have the following (for the first condition):
So, say we have Subtract (2) as the
operationSince the previous result is
00000anything AND'd with it will be zero. So your condition will always evaluate totruesince0 == 0.If we just switch this to OR, then we have the following:
Now, say we have
Add (1)as theoperationSo,
1 & 17 => 1and thus your final condition is(1 & ( 1 | 16 ) ) == ( 1 | 16 )=>1 & 17 == 17=>1 == 17=>false(still false!)So what you actually want is:
This becomes
((1 | 1 | 16 ) & ( 1 | 16 )) == ( 1 | 16 )=>( 17 & 17 ) == 17=>17 == 17==trueThis is obviously not readable, so you should opt for extracting this into a method (as suggested). But it's still important to understand why your condition is incorrect.
重复：
请参阅如何在 C# 中比较标志？
C# 4 引入了 Enum.HasFlags() 方法。
但是，请参阅 是什么让 Enum.HasFlag 如此缓慢？ 如果性能是一个问题。
Duplicate:
See How to Compare Flags in C#?
C# 4 introduces the Enum.HasFlags() method.
However, see What is it that makes Enum.HasFlag so slow? if perfomance is an issue.
你的操作失败的原因是你的表达式错误。
(Operations.add & Operations.eval)始终为零。第一次比较的左侧和右侧始终为零。试试这个 - 我怀疑这就是你想要的：The reason your operation is failing is because you have the wrong expression.
(Operations.add & Operations.eval)is always zero. Left and right side of your first comparison are both always zero. Try this instead - I suspect it's what you were after:试试这个：
Try this: