在 Windows Forms 中检查 DrawItemState 等价性的条件表达式是什么意思?
What is the meaning of the conditional expression for checking DrawItemState equivalence in Windows Forms?
我在阅读 DrawItemState documentation 时遇到了以下代码片段:
if ((e.State & DrawItemState.Selected) == DrawItemState.Selected )
brush = SystemBrushes.HighlightText;
文档中给出了以下解释
Since state can be a combination (bit-flag) of enum values, you can't
use "==" to compare them
但是我仍然不明白这个条件表达式与下面显示的片段有何不同:
if (e.State == DrawItemState.Selected )
brush = SystemBrushes.HighlightText;
此外,我不明白按位与 &
运算符有何不同,以及为什么它甚至包含在条件表达式中。
由于枚举的基础值,您可以设置多个值(bitwise combination):
var state = DrawItemState.Disabled | DrawItemState.Grayed;
这意味着这些都将 return false
:
Console.WriteLine(state == DrawItemState.Disabled); // false
Console.WriteLine(state == DrawItemState.Grayed); // false
测试值的一种方法是使用按位 "and" 运算符:
Console.WriteLine((state & DrawItemState.Grayed) == DrawItemState.Grayed); // true
基本上,在我的示例中,state
设置了两个位 - "Grayed" 和 "Disabled"。通过对 "Grayed" 的值使用按位 "and" 运算符,结果也是 "Grayed".
的值
0 0 0 0 0 0 0 0 1 1 0 // binary value of disabled and grayed
0 0 0 0 0 0 0 0 0 1 0 // binary value of grayed
0 0 0 0 0 0 0 0 0 1 0 // result is same value as grayed
您也可以测试多个标志:
Console.WriteLine((state & (DrawItemState.Grayed | DrawItemState.Disabled))
== (DrawItemState.Grayed | DrawItemState.Disabled)); // true
0 0 0 0 0 0 0 0 1 1 0 // binary value of disabled and grayed
0 0 0 0 0 0 0 0 1 1 0 // binary value of disabled and grayed
0 0 0 0 0 0 0 0 1 1 0 // result is same value as disabled and grayed
就个人而言,我认为使用 HasFlag
:
来测试标志值更容易
Console.WriteLine(state.HasFlag(DrawItemState.Grayed)); // true
请注意,如果只有一个值,则 ==
会起作用,但对于支持按位组合的 "flags" 枚举,这永远无法保证。
var state = DrawItemState.Grayed;
Console.WriteLine(state == DrawItemState.Grayed); // true
我在阅读 DrawItemState documentation 时遇到了以下代码片段:
if ((e.State & DrawItemState.Selected) == DrawItemState.Selected )
brush = SystemBrushes.HighlightText;
文档中给出了以下解释
Since state can be a combination (bit-flag) of enum values, you can't use "==" to compare them
但是我仍然不明白这个条件表达式与下面显示的片段有何不同:
if (e.State == DrawItemState.Selected )
brush = SystemBrushes.HighlightText;
此外,我不明白按位与 &
运算符有何不同,以及为什么它甚至包含在条件表达式中。
由于枚举的基础值,您可以设置多个值(bitwise combination):
var state = DrawItemState.Disabled | DrawItemState.Grayed;
这意味着这些都将 return false
:
Console.WriteLine(state == DrawItemState.Disabled); // false
Console.WriteLine(state == DrawItemState.Grayed); // false
测试值的一种方法是使用按位 "and" 运算符:
Console.WriteLine((state & DrawItemState.Grayed) == DrawItemState.Grayed); // true
基本上,在我的示例中,state
设置了两个位 - "Grayed" 和 "Disabled"。通过对 "Grayed" 的值使用按位 "and" 运算符,结果也是 "Grayed".
0 0 0 0 0 0 0 0 1 1 0 // binary value of disabled and grayed
0 0 0 0 0 0 0 0 0 1 0 // binary value of grayed
0 0 0 0 0 0 0 0 0 1 0 // result is same value as grayed
您也可以测试多个标志:
Console.WriteLine((state & (DrawItemState.Grayed | DrawItemState.Disabled))
== (DrawItemState.Grayed | DrawItemState.Disabled)); // true
0 0 0 0 0 0 0 0 1 1 0 // binary value of disabled and grayed
0 0 0 0 0 0 0 0 1 1 0 // binary value of disabled and grayed
0 0 0 0 0 0 0 0 1 1 0 // result is same value as disabled and grayed
就个人而言,我认为使用 HasFlag
:
Console.WriteLine(state.HasFlag(DrawItemState.Grayed)); // true
请注意,如果只有一个值,则 ==
会起作用,但对于支持按位组合的 "flags" 枚举,这永远无法保证。
var state = DrawItemState.Grayed;
Console.WriteLine(state == DrawItemState.Grayed); // true