C++ (1U << X) 有什么作用？
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What does (1U << X) do?
提问by Karl von Moor
I found this piece of code:
我找到了这段代码：
enum
{
IsDynamic = (1U << 0), // ...
IsSharable = (1U << 1), // ...
IsStrong = (1U << 2) // ...
};
What does the (1U << X)do?
有什么作用(1U << X)？
回答by Philippe Leybaert
It sets bitmasks:
它设置位掩码：
1U << 0 = 1
1U << 1 = 2
1U << 2 = 4
etc...
What happens is 1U (unsigned value 1) is shifted to the left by x bits.
发生的情况是 1U（无符号值 1）向左移动 x 位。
The code you posted is equivalent to:
您发布的代码相当于：
enum
{
IsDynamic = 1U, // binary: 00000000000000000000000000000001
IsSharable = 2U, // binary: 00000000000000000000000000000010
IsStrong = 4U // binary: 00000000000000000000000000000100
}
回答by Hamish Grubijan
Bit shift. Instead of saying a = 1, b = 2, c = 4 they shift the bits. The idea is to pack many flags into one integer (or long).
位移。他们不是说 a = 1、b = 2、c = 4，而是移位了这些位。这个想法是将许多标志打包成一个整数（或长整数）。
This is actually a very clean approach.
这实际上是一种非常干净的方法。
回答by Ben Gottlieb
<< is the bitshift operator. It will take the bits in the left side and shift them by an amount specified by the right side. For example:
<< 是位移运算符。它将获取左侧的位并将它们移位右侧指定的数量。例如：
1 << 1 -> 0b0001 << 1 => 0b0010
1 << 2 -> 0b0001 << 2 => 0b0100
etc.
等等。
回答by Mark Ransom
1Uis an unsigned value with the single bit 0 set, and all the other bits cleared. The <<operator means "shift to the left". 1U << 0means create a value with bit 0 set; 1U << 1means create a value with bit 1 set; etc.
1U是一个无符号值，设置了单个位 0，并且所有其他位都被清除。该<<运营商表示“左移”。 1U << 0表示创建一个设置为位 0 的值；1U << 1表示创建一个设置位 1 的值；等等。
回答by Bill Bingham
That snippet
那个片段
enum
{
IsDynamic = (1U << 0), // ...
IsSharable = (1U << 1), // ...
IsStrong = (1U << 2) // ...
}
declares an enumeration with values which are powers of 2. To be used presumably as masks on a value which contains multiple flags.
声明一个枚举，其值为 2 的幂。大概用作包含多个标志的值的掩码。
So for example a value representing something that IsDynamic and IsSharable is
例如，表示 IsDynamic 和 IsSharable 的值是
unsigned value = IsDynamic | IsSharable; // could use + as well
And to test if the value IsStrong
并测试值是否为 IsStrong
if (value & IsStrong) { ... }
