blocks|key|4511923|text|给定ch+=+18|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4511924|(ch+&+16)+=+16
(ch+&+8)+=+0
(ch+&+2)+=+2|code-block|syntax|javascript|4511925|看起来是对的。18在二进制中是什么？16+%7C+2或10010。|4511926|需要说明的是，&是一个按位运算符。但是，如果两个操作数都为真，则&&返回true。|4511927|entityMap^0|2|7|0|0|I|6|P|5|0|7|1|W|2|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|Z|8|@$9|10|A|11|B|C]|$9|12|A|13|B|C]]|D|@]|E|$]]|$1|O|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|P|$]]

Given <code>ch = 18</code>

<pre><code>(ch &amp; 16) = 16
(ch &amp; 8) = 0
(ch &amp; 2) = 2
</code></pre>

seems correct. What's 18 in binary ? <code>16 | 2</code> or <code>10010</code> in binary.

To be clear, <code>&amp;</code> is a bitwise operator. However <code>&amp;&amp;</code> returns true if both operands are true.

blocks|key|4602307|text|&是bitwise+operator，并且：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4602308|18+=+10010
16+=+10000
----------
16+=+10000


18+=+10010
+8+=+01000
----------
+0+=+00000|code-block|syntax|javascript|4602309|因此if将检查第五位是1还是0。|4602310|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/tutorial/java/nutsandbolts/op3.html^0|0|1|2|G|0|0|0|2|2|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@$9|X|A|Y|1|Z]]|E|$]]|$1|F|3|G|5|H|7|10|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|11|8|@$9|12|A|13|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

<code>&amp;</code> is the <a href="http://docs.oracle.com/javase/tutorial/java/nutsandbolts/op3.html" rel="nofollow noreferrer">bitwise operator</a> AND:

<pre><code>18 = 10010
16 = 10000
----------
16 = 10000


18 = 10010
 8 = 01000
----------
 0 = 00000
</code></pre>

So the <code>if</code> will check if the fifth bit is 1 or 0.

blocks|key|4602339|text|单与号是一个按位AND运算符。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4602340|如果值是十六进制的，您可能会发现更容易“获取”。|4602341|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

Single ampersand is a bitwise AND operator.

You might find it easier to "get" if the values were in hex.

blocks|key|4602389|text|The+single+&+it's+a+bitwise+AND.它是对数字的单个位执行的and操作。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4602390|考虑一下你的短片的一种可能的位表示：|4602391|10011011+&+++:+screendata[pos]
00010000+=+++:+16
----------
10010000|code-block|syntax|javascript|4602392|特别是这一行：|4602393|if+((ch+&+16)+!=+0)+{|4602394|检查数字的第5位(2%5E+(5+-1))是否设置为1(不同于0)。|4602395|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Bitwise_operation^0|B|1|S|3|K|7|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@$9|15|A|16|1|17]]|E|$]]|$1|F|3|G|5|6|7|18|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|19|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|1A|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|1B|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|1C|8|@]|D|@]|E|$]]|$1|S|3|-4|5|6|7|1D|8|@]|D|@]|E|$]]]|T|$U|$5|V|W|X|E|$Y|Z]]]]

The single <code>&amp;</code> it's a <a href="http://en.wikipedia.org/wiki/Bitwise_operation" rel="nofollow">bitwise</a> <code>AND</code>. It's an and operation performed on individual bit of your number.

Consider a possible bit representation of your short:

<pre><code>10011011 &amp; : screendata[pos]
00010000 = : 16
----------
10010000
</code></pre>

Specifically this line:

<pre><code>if ((ch &amp; 16) != 0) {
</code></pre>

check if the 5-th bit (2^ (5 -1)) of your number is set to 1 (different from 0).

blocks|key|4602422|text|按位and运算符&对两个整数的每一位执行"and“以确定结果。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4602423|所以：|4602424|18+=+0001+0010|4602425|16+=+0001+0000|4602426|18&16+=+0001+0000|4602427|entityMap^0|8|1|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|S|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|T|8|@]|D|@]|E|$]]|$1|J|3|K|5|6|7|U|8|@]|D|@]|E|$]]|$1|L|3|M|5|6|7|V|8|@]|D|@]|E|$]]|$1|N|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|O|$]]

The bitwise and operator <code>&amp;</code> performs an "and" on each bit of the two integers to determine the result.

So:

18 = 0001 0010 
16 = 0001 0000 
18&amp;16 = 0001 0000

blocks|key|2874879|text|这不是一个布尔and，它总是&&；相反，它是一个按位and。它正在检查是否在ch中设置了右起第五位。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2874880|ch+=+18++//++++++ch+=+0b00010100
ch+&+16++//++++++16+=+0b00010000
+++++++++//+ch+&+16+=+0b00010000+!=+0
ch+&+8+++//+++++++8+=+0b00001000
+++++++++//+ch+&++8+=+0b00000000+==+0
ch+&+2+++//+++++++2+=+0b00000010
+++++++++//+ch+&++2+=+0b00000010+!=+0|code-block|syntax|javascript|2874881|entityMap^0|E|2|12|2|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

That's not a Boolean and, which is always <code>&amp;&amp;</code>; instead it's a bitwise and. It's checking to see if the 5th bit from the right is set in <code>ch</code>.

<pre><code>ch = 18 // ch = 0b00010100
ch &amp; 16 // 16 = 0b00010000
 // ch &amp; 16 = 0b00010000 != 0
ch &amp; 8 // 8 = 0b00001000
 // ch &amp; 8 = 0b00000000 == 0
ch &amp; 2 // 2 = 0b00000010
 // ch &amp; 2 = 0b00000010 != 0
</code></pre>

blocks|key|4512114|text|&是按位运算符AND。您可以在http://docs.oracle.com/javase/tutorial/java/nutsandbolts/op3.html中查看详细信息|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|4512115|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/tutorial/java/nutsandbolts/op3.html^0|F|1T|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

&amp; is bitwise operator AND. You can see detail in <a href="http://docs.oracle.com/javase/tutorial/java/nutsandbolts/op3.html" rel="nofollow">http://docs.oracle.com/javase/tutorial/java/nutsandbolts/op3.html</a>

I am reading a tutorial regarding a Java pacman game. 

Here is the code in question.

<pre><code> if (pacmanx % blocksize == 0 &amp;&amp; pacmany % blocksize == 0) {
 pos = // integer 
 ch = screendata[pos];

 if ((ch &amp; 16) != 0) { // do not understand this.
 screendata[pos] = (short)(ch &amp; 15);
 ...
 }
</code></pre>

I am not really understanding the single &amp;. I understand this operand checks both sides of an if statement, or is a bitwise operator. However, per the tests below, it doesn't seem to be either:

<pre><code>if I was to test (ch = 18):
(ch &amp; 16) = 16
(ch &amp; 8) = 0
(ch &amp; 2) = 2
</code></pre>

thanks

having trouble understanding code, specifically the &

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我正在读一篇关于Java pacman游戏的教程。下面是有问题的代码。   if (pacmanx % blocksize == 0 && pacmany % blocksize == 0) {        pos = // integer                 ch = screendata[pos]; ...

问理解代码有困难，特别是&
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问理解代码有困难，特别是&EN