什么是这个声明的含义是什么？ [英] What is the meaning of this declaration?

问题描述

我不是在C / C ++的专家。

I am not a specialist in C/C++.

今天我发现了这个声明：

I found this declaration today:

typedef NS_OPTIONS(NSUInteger, PKRevealControllerType)
{
    PKRevealControllerTypeNone  = 0,
    PKRevealControllerTypeLeft  = 1 << 0,
    PKRevealControllerTypeRight = 1 << 1,
    PKRevealControllerTypeBoth = (PKRevealControllerTypeLeft | PKRevealControllerTypeRight)
};

你们可以翻译一下值每个值将有？

Can you guys translate what values every value will have?

推荐答案

算子的＆LT;＆LT; 终止的按位的左移操作符。移所有位向左的指定次数：（算术左移和储备符号位）

opertor << is bitwise left shift operator. Shift all the bits to left a specified number of times: (arithmetic left shift and reserves sign bit)

m << n

移 M 的所有位向左一个 N 的次数。 （的通知由两个单班==乘的）。

Shift all the bits of m to left a n number of times. (notice one shift == multiply by two).

1 LT;＆LT; 0 手段的没有的转变所以它等于 1 而已。

1 << 0 means no shift so its equals to 1 only.

1 LT;＆LT; 1 手段的有一个的转变所以它等于 1 * 2 = 2只。

1 << 1 means one shift so its equals to 1*2 = 2 only.

我跟一个字节解释：一个在一个字节是这样的：

I explain with one byte: one in one byte is like:

 MSB
+----+----+----+---+---+---+---+---+
|  0 |  0 | 0  | 0 | 0 | 0 | 0 | 1 |       1   
+----+----+----+---+---+---+---+---+
  7     6   5    4   3   2   1 / 0  
  |                           /           1 << 1
  |                          | 
  ▼                          ▼
+----+----+----+---+---+---+---+---+
|  0 |  0 | 0  | 0 | 0 | 0 | 1 | 0 |       2      
+----+----+----+---+---+---+---+---+ 
   7    6   5    4   3   2   1   0

而 1 LT;＆LT; 0 做什么，但它一样的人物之一。 （注意第7位被复制到preserve号）

Whereas 1 << 0 do nothing but its like figure one. (notice 7th bit is copied to preserve sign)

或接线员：逐位或

 MSB                            PKRevealControllerTypeLeft
+----+----+----+---+---+---+---+---+
|  0 |  0 | 0  | 0 | 0 | 0 | 0 | 1 |  == 1
+----+----+----+---+---+---+---+---+
   7    6   5    4   3   2   1   0
   |    |    |   |   |   |   |   |      OR
 MSB                               PKRevealControllerTypeRight
+----+----+----+---+---+---+---+---+
|  0 |  0 | 0  | 0 | 0 | 0 | 1 | 0 |   == 2
+----+----+----+---+---+---+---+---+
   7    6   5    4   3   2   1   0

 = 

 MSB                    PKRevealControllerTypeBoth
+----+----+----+---+---+---+---+---+
|  0 |  0 | 0  | 0 | 0 | 0 | 1 | 1 |   == 3
+----+----+----+---+---+---+---+---+
   7    6   5    4   3   2   1   0  

| 是位聪明的运营商。下面code将其或 1 | 2 == 3

| is bit wise operator. in below code it or 1 | 2 == 3

PKRevealControllerTypeNone  = 0,             //  is Zero
PKRevealControllerTypeLeft  = 1 << 0,        //  one 
PKRevealControllerTypeRight = 1 << 1,        //  two
PKRevealControllerTypeBoth = (PKRevealControllerTypeLeft |    
                             PKRevealControllerTypeRight)  // three

有没有更多的技术理由来这样的初始化值，定义​​这样使事情排队很好地阅读这样的回答：的定义了一种（1 LT;℃下）

There is not more technical reason to initialized values like this, defining like that makes things line up nicely read this answer:define SOMETHING (1 << 0)

编译器优化其转换更简单的像：（我不知道对于第三个，但我认为编译器将优化太的）

compiler optimization convert them in simpler for like: (I am not sure for third one, but i think compiler will optimize that too)

PKRevealControllerTypeNone  = 0,     //  is Zero
PKRevealControllerTypeLeft  = 1,     //  one 
PKRevealControllerTypeRight = 2,     //  two
PKRevealControllerTypeBoth  = 3,     // Three

编辑： @thanks耕种。
阅读这个答案<一个href=\"http://stackoverflow.com/questions/8956364/app-states-with-bool-flags?answertab=votes#8956606\">App与BOOL标志美国显示你有使用逐位运算符声明的有效性。

@thanks to Till. read this answer App States with BOOL flags show the usefulness of declarations you got using bit wise operators.

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