blocks|key|1397400|text|我会使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1397401|if+((value+&+(1L+<<+x))+!=+0)
{
+++//+The+bit+was+set
}|code-block|syntax|javascript|1397402|(您可能可以使用较少的括号，但我从来不记得按位操作的优先顺序。)|1397403|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I'd use:

<pre><code>if ((value &amp; (1L &lt;&lt; x)) != 0)
{
 // The bit was set
}
</code></pre>

(You may be able to get away with fewer brackets, but I never remember the precedence of bitwise operations.)

blocks|key|1068948|text|另一种选择：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1068949|if+(BigInteger.valueOf(value).testBit(x))+{
++++//+...
}|code-block|syntax|javascript|1068950|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Another alternative:

<pre><code>if (BigInteger.valueOf(value).testBit(x)) {
 // ...
}
</code></pre>

blocks|key|1068987|text|我想知道是不是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1068988|++if+(((value+>>>+x)+&+1)+!=+0)+{

++}|code-block|syntax|javascript|1068989|..更好是因为值是否很长并不重要，或者更糟糕是因为它不那么明显。|offset|length|style|BOLD|1068990|汤姆·霍廷-7月7日14:16的抢断线|1068991|entityMap^0|0|0|0|W|0|0|F|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|V|8|@$I|W|J|X|K|L]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|P|$]]

I wonder if:

<pre><code> if (((value &gt;&gt;&gt; x) &amp; 1) != 0) {

 }
</code></pre>

.. is better because it doesn't matter whether value is long or not, or if its worse because it's less obvious.

Tom Hawtin - tackline Jul 7 at 14:16

blocks|key|1399340|text|您还可以使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1399341|bool+isSet+=+((value>>x)+&+1)+!=+0;|code-block|syntax|javascript|1399342|编辑："(value>>x)+&+1“和"value+&+(1<<x)”之间的区别取决于当x大于"value“类型的大小(在您的例子中为32)时的行为。|offset|length|style|CODE|1399343|在这种情况下，使用"(value>>x)+&+1“会得到值的符号，而使用"value+&+(1<<x)”会得到0(如果x太大，有时获取位符号是很有用的)。|1399344|如果您希望在这种情况下使用0，则可以使用">>>“运算符，如果是">>”|1399345|因此，"((value>>>x)+&+1)+!=+0“和"(value+&+(1<<x))+!=+0”是完全等价的|1399346|entityMap^0|0|0|4|E|L|E|0|A|E|11|E|0|L|3|X|2|0|4|M|T|L|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@$I|X|J|Y|K|L]|$I|Z|J|10|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|11|8|@$I|12|J|13|K|L]|$I|14|J|15|K|L]]|9|@]|A|$]]|$1|O|3|P|5|6|7|16|8|@$I|17|J|18|K|L]|$I|19|J|1A|K|L]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1B|8|@$I|1C|J|1D|K|L]|$I|1E|J|1F|K|L]]|9|@]|A|$]]|$1|S|3|-4|5|6|7|1G|8|@]|9|@]|A|$]]]|T|$]]

You can also use

<pre><code>bool isSet = ((value&gt;&gt;x) &amp; 1) != 0;
</code></pre>

EDIT: the difference between "<code>(value&gt;&gt;x) &amp; 1</code>" and "<code>value &amp; (1&lt;&lt;x)</code>" relies on the behavior when x is greater than the size of the type of "value" (32 in your case).

In that particular case, with "<code>(value&gt;&gt;x) &amp; 1</code>" you will have the sign of value, whereas you get a 0 with "<code>value &amp; (1&lt;&lt;x)</code>" (it is sometimes useful to get the bit sign if x is too large).

If you prefer to have a 0 in that case, you can use the "<code>&gt;&gt;&gt;</code>" operator, instead if "<code>&gt;&gt;</code>"

So, "<code>((value&gt;&gt;&gt;x) &amp; 1) != 0</code>" and "<code>(value &amp; (1&lt;&lt;x)) != 0</code>" are completely equivalent

blocks|key|1397414|text|对于nth+LSB+(最低有效位)，应执行以下操作：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1397415|boolean+isSet+=+(value+&+(1+<<+n))+!=+0;|code-block|syntax|javascript|1397416|entityMap^0|2|1|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

For the <code>n</code>th LSB (least significant bit), the following should work:

<pre><code>boolean isSet = (value &amp; (1 &lt;&lt; n)) != 0;
</code></pre>

blocks|key|4286065|text|您可能希望查看BitSet：http://java.sun.com/javase/6/docs/api/java/util/BitSet.html|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|4286066|entityMap|0|LINK|mutability|MUTABLE|url|http://java.sun.com/javase/6/docs/api/java/util/BitSet.html^0|E|1N|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]]]]

You might want to check out BitSet: <a href="http://java.sun.com/javase/6/docs/api/java/util/BitSet.html" rel="noreferrer">http://java.sun.com/javase/6/docs/api/java/util/BitSet.html</a>

blocks|key|98035|text|对x执行Bit+shifting操作，并检查最低位。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|98036|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Bitwise_operation#Arithmetic_shift^0|4|C|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]]]]

<a href="http://en.wikipedia.org/wiki/Bitwise_operation#Arithmetic_shift" rel="nofollow noreferrer">Bit shifting</a> right by x and checking the lowest bit.

blocks|key|97398|text|在Java中，以下代码运行良好：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|97399|if+(value+<<+~x+<+0)+{
+++//+xth+bit+set
}+else+{
+++//+xth+bit+not+set
}|code-block|syntax|javascript|97400|value和x可以是int或long+(不需要相同)。|offset|length|style|CODE|97401|对非程序员的警告：前面的表达式在Java语言中有效，因为在Java语言中，位移位运算符仅适用于右侧操作数的5位(如果是long，则为6位)。这会将表达式隐式转换为value+<<+(~x+&+31)+(如果value为long，则为value+<<+(~x+&+63)+)。|BOLD|97402|+javascript+：它也适用于javascript(像java一样，只应用移位计数的最低5位)。在javascript中，任何number都是32位的。|97403|特别是在C中，负移位计数会调用未定义的行为，因此此测试不一定有效(尽管它可能有效，这取决于您的编译器/处理器的特定组合)。|97404|它怎麽工作?|97405|这个答案的巧妙之处在于整数的符号位非常容易读取:当设置该位时，该值为负数；如果未设置，则该值为零或正数。|97406|所以整个想法是将第X位精确地移到符号位中。这意味着31+-+x向左移动(如果value是64位宽，则为63+-+x+)。|97407|在java+(以及其他语言)中，~运算符计算bitwise+NOT+operation，在算术上等于-x+-+1+(无论x有多宽)。|97408|此外，java+<<运算符仅采用右侧操作数的最低有效5(或6)位(5或6取决于左侧操作数的宽度:对于int，则为5；对于long，则为6)。从算术上讲，这与除以32+(或64)的余数相同。|97409|这就是(-x+-+1)+%25+32+=+31+-+x+(或者(-x+-+1)+%25+64+=+63+-+x，对于64位宽的value)。|97410|entityMap|0|LINK|mutability|MUTABLE|url|https://en.wikipedia.org/wiki/Bitwise_operation#NOT^0|0|0|0|5|6|1|A|3|E|4|0|2|6|1N|4|29|I|2V|5|31|4|38|I|0|0|C|1U|6|0|0|5|0|0|0|P|6|12|5|1F|6|0|G|1|U|3|1E|6|1O|1|M|L|0|0|8|2|1E|3|1O|4|0|3|D|J|6|T|D|19|6|1N|5|0^^$0|@$1|2|3|4|5|6|7|1D|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|1E|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|1F|8|@$I|1G|J|1H|K|L]|$I|1I|J|1J|K|L]|$I|1K|J|1L|K|L]|$I|1M|J|1N|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|1O|8|@$I|1P|J|1Q|K|O]|$I|1R|J|1S|K|L]|$I|1T|J|1U|K|L]|$I|1V|J|1W|K|L]|$I|1X|J|1Y|K|L]|$I|1Z|J|20|K|L]]|9|@]|A|$]]|$1|P|3|Q|5|6|7|21|8|@$I|22|J|23|K|O]|$I|24|J|25|K|L]]|9|@]|A|$]]|$1|R|3|S|5|6|7|26|8|@$I|27|J|28|K|O]]|9|@]|A|$]]|$1|T|3|U|5|6|7|29|8|@]|9|@]|A|$]]|$1|V|3|W|5|6|7|2A|8|@]|9|@]|A|$]]|$1|X|3|Y|5|6|7|2B|8|@$I|2C|J|2D|K|L]|$I|2E|J|2F|K|L]|$I|2G|J|2H|K|L]]|9|@]|A|$]]|$1|Z|3|10|5|6|7|2I|8|@$I|2J|J|2K|K|L]|$I|2L|J|2M|K|L]|$I|2N|J|2O|K|L]|$I|2P|J|2Q|K|L]]|9|@$I|2R|J|2S|1|2T]]|A|$]]|$1|11|3|12|5|6|7|2U|8|@$I|2V|J|2W|K|L]|$I|2X|J|2Y|K|L]|$I|2Z|J|30|K|L]]|9|@]|A|$]]|$1|13|3|14|5|6|7|31|8|@$I|32|J|33|K|L]|$I|34|J|35|K|L]|$I|36|J|37|K|L]|$I|38|J|39|K|L]|$I|3A|J|3B|K|L]]|9|@]|A|$]]|$1|15|3|-4|5|6|7|3C|8|@]|9|@]|A|$]]]|16|$17|$5|18|19|1A|A|$1B|1C]]]]

In Java the following works fine:
<pre><code>if (value &lt;&lt; ~x &lt; 0) {
 // xth bit set
} else {
 // xth bit not set
}
</code></pre>
<code>value</code> and <code>x</code> can be <code>int</code> or <code>long</code> (and don't need to be the same).
Word of caution for non-Java programmers: the preceding expression works in Java because in that language the bit shift operators apply only to the 5 (or 6, in case of <code>long</code>) lowest bits of the right hand side operand. This implicitly translates the expression to <code>value &lt;&lt; (~x &amp; 31)</code> (or <code>value &lt;&lt; (~x &amp; 63)</code> if <code>value</code> is <code>long</code>).
Javascript: it also works in javascript (like java, only the lowest 5 bits of shift count are applied). In javascript any <code>number</code> is 32-bit.
Particularly in C, negative shift count invokes undefined behavior, so this test won't necessarily work (though it may, depending on your particular combination of compiler/processor).
<h3>How does it work?</h3>
The cleverness of this answer lies in that the sign bit of an integer is very easy to read: when that bit is set, then the value is negative; if not set, then the value is either zero or positive.
So the whole idea is to shift the xth bit exactly into the sign bit. That means a displacement of <code>31 - x</code> to the left (or <code>63 - x</code>, if <code>value</code> is 64 bits wide).
In java (and other languages as well), the <code>~</code> operator computes the <a href="https://en.wikipedia.org/wiki/Bitwise_operation#NOT" rel="nofollow noreferrer">bitwise <code>NOT</code> operation</a>, which arithmetically equals to <code>-x - 1</code> (no matter how wide <code>x</code> is).
Also the java <code>&lt;&lt;</code> operator takes only the least significant 5 (or 6) bits of the right-hand side operand (whether 5 or 6 depends on how wide the left-hand side operand is: for <code>int</code> then 5; for <code>long</code> then 6). Arithmetically this is the same as the remainder of division by 32 (or 64).
And that is <code>(-x - 1) % 32</code> = <code>31 - x</code> (or <code>(-x - 1) % 64</code> = <code>63 - x</code>, for 64 bit wide <code>value</code>).

blocks|key|1397429|text|2%5Ex位的值是"variable+&+(1+<<+x)“|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1397430|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

The value of the 2^x bit is "variable &amp; (1 &lt;&lt; x)"

blocks|key|4286090|text|我的贡献-忽略前一个|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4286091|public+class+TestBits+{+

++++public+static+void+main(String[]+args)+{+

++++++++byte+bit1+=+0b00000001;+++++
++++++++byte+bit2+=+0b00000010;
++++++++byte+bit3+=+0b00000100;
++++++++byte+bit4+=+0b00001000;
++++++++byte+bit5+=+0b00010000;
++++++++byte+bit6+=+0b00100000;
++++++++byte+bit7+=+0b01000000;

++++++++byte+myValue+=+9;++++++++++++++++++++++++//+any+value

++++++++if+(((myValue+>>>+3)+&+bit1+)+!=+0)+{++++//++shift+3+to+test+bit4
++++++++++++System.out.println("+ON+");+
++++++++}
++++}+
}|code-block|syntax|javascript|4286092|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

My contribution - ignore previous one

<pre><code>public class TestBits { 

 public static void main(String[] args) { 

 byte bit1 = 0b00000001; 
 byte bit2 = 0b00000010;
 byte bit3 = 0b00000100;
 byte bit4 = 0b00001000;
 byte bit5 = 0b00010000;
 byte bit6 = 0b00100000;
 byte bit7 = 0b01000000;

 byte myValue = 9; // any value

 if (((myValue &gt;&gt;&gt; 3) &amp; bit1 ) != 0) { // shift 3 to test bit4
 System.out.println(" ON "); 
 }
 } 
}
</code></pre>

blocks|key|97282|text|如果有人对按位运算符不太熟悉，那么可以尝试下面的代码来编程决定它。有两种方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|97283|1)使用java语言功能获取二进制格式的字符串，然后检查特定位置的字符。|97284|2)继续除以2，确定某一位置的位值。|97285|public+static+void+main(String[]+args)+{
++++Integer+n+=1000;
++++String+binaryFormat+=++Integer.toString(n,+2);
++++int+binaryFormatLength+=+binaryFormat.length();
++++System.out.println("binaryFormat="%2BbinaryFormat);
++++for(int+i+=+1;i<10;i%2B%2B){
++++++++System.out.println("isBitSet("%2Bn%2B","%2Bi%2B")"%2BisBitSet(n,i));
++++++++System.out.println((binaryFormatLength>=i+&&+binaryFormat.charAt(binaryFormatLength-i)=='1'));
++++}

}

public+static+boolean+isBitSet(int+number,+int+position){
++++int+currPos+=1;
++++int+temp+=+number;
++++while(number!=0+&&+currPos<=+position){
++++++++if(temp%252+==+1+&&+currPos+==+position)
++++++++++++return+true;
++++++++else{
++++++++++++temp+=+temp/2;
++++++++++++currPos+%2B%2B;
++++++++}
++++}
++++return+false;
}|code-block|syntax|javascript|97286|输出|97287|binaryFormat=1111101000
isBitSet(1000,1)false
false
isBitSet(1000,2)false
false
isBitSet(1000,3)false
false
isBitSet(1000,4)true
true
isBitSet(1000,5)false
false
isBitSet(1000,6)true
true
isBitSet(1000,7)true
true
isBitSet(1000,8)true
true
isBitSet(1000,9)true
true|97288|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|S|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|T|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|H|7|V|8|@]|9|@]|A|$I|J]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

If someone is not very comfortable with bitwise operators, then below code can be tried to programatically decide it. There are two ways.

1) Use java language functionality to get the binary format string and then check character at specific position

2) Keep dividing by 2 and decide the bit value at certain position. 

<pre><code>public static void main(String[] args) {
 Integer n =1000;
 String binaryFormat = Integer.toString(n, 2);
 int binaryFormatLength = binaryFormat.length();
 System.out.println("binaryFormat="+binaryFormat);
 for(int i = 1;i&lt;10;i++){
 System.out.println("isBitSet("+n+","+i+")"+isBitSet(n,i));
 System.out.println((binaryFormatLength&gt;=i &amp;&amp; binaryFormat.charAt(binaryFormatLength-i)=='1'));
 }

}

public static boolean isBitSet(int number, int position){
 int currPos =1;
 int temp = number;
 while(number!=0 &amp;&amp; currPos&lt;= position){
 if(temp%2 == 1 &amp;&amp; currPos == position)
 return true;
 else{
 temp = temp/2;
 currPos ++;
 }
 }
 return false;
}
</code></pre>

Output

<pre><code>binaryFormat=1111101000
isBitSet(1000,1)false
false
isBitSet(1000,2)false
false
isBitSet(1000,3)false
false
isBitSet(1000,4)true
true
isBitSet(1000,5)false
false
isBitSet(1000,6)true
true
isBitSet(1000,7)true
true
isBitSet(1000,8)true
true
isBitSet(1000,9)true
true
</code></pre>

blocks|key|1399416|text|声明一个temp+int并使其等于原始int。然后将temp+>>移位x次，以便要检查的位位于最后位置。然后执行temp+&+0xf以删除前面的位。现在剩下最后一点了。最后做if+(y+&+1+==+0)，如果最后一位是1，它应该等于0，否则将等于1。它要么是那个，要么是if+(y%2B0x1+==+0)...不太确定。到处闲逛，看看|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1399417|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

declare a temp int and make it equal the original. 
then shift temp >> x times, so that the bit you want to check is at the last position. then do temp &amp; 0xf to drop the preceding bits. Now left with last bit. Finally do if (y &amp; 1 == 0), if last bit is a 1, that should equal 0, else will equal 1. Its either that or if (y+0x1 == 0)... not too sure. fool around and see

blocks|key|97303|text|我编写了一个小的静态类，它做了一些位操作。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|97304|public+final+class+Bitfield+{

++private+Bitfield()+{}

++//+********************************************************************
++//+*+TEST
++//+********************************************************************

++public+static+boolean+testBit(final+int+pos,+final+int+bitfield)+{
++++++return+(bitfield+&+(1+<<+pos))+==+(1+<<+pos);
++}

++public+static+boolean+testNum(final+int+num,+final+int+bitfield)+{
++++++return+(bitfield+&+num)+==+num;
++}

++//+********************************************************************
++//+*+SET
++//+********************************************************************

++public+static+int+setBit(final+int+pos,+final+int+bitfield)+{
+++++return+bitfield+%7C+(1+<<+pos);
++}

++public+static+int+addNum(final+int+number,+final+int+bitfield)+{
++++++return+bitfield+%7C+number;
++}

++//+********************************************************************
++//+*+CLEAR
++//+********************************************************************

++public+static+int+clearBit(final+int+pos,+final+int+bitfield)+{
++++++return+bitfield+%5E+(1+<<+pos);
++}

++public+static+int+clearNum(final+int+num,+final+int+bitfield)+{
++++++return+bitfield+%5E+num;
++}

++}|code-block|syntax|javascript|97305|如果有什么问题，请给我写封电子邮件。|97306|很好的编程！|97307|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I coded a little static class which is doing some of the bit operation stuff.

<pre><code>public final class Bitfield {

 private Bitfield() {}

 // ********************************************************************
 // * TEST
 // ********************************************************************

 public static boolean testBit(final int pos, final int bitfield) {
 return (bitfield &amp; (1 &lt;&lt; pos)) == (1 &lt;&lt; pos);
 }

 public static boolean testNum(final int num, final int bitfield) {
 return (bitfield &amp; num) == num;
 }

 // ********************************************************************
 // * SET
 // ********************************************************************

 public static int setBit(final int pos, final int bitfield) {
 return bitfield | (1 &lt;&lt; pos);
 }

 public static int addNum(final int number, final int bitfield) {
 return bitfield | number;
 }

 // ********************************************************************
 // * CLEAR
 // ********************************************************************

 public static int clearBit(final int pos, final int bitfield) {
 return bitfield ^ (1 &lt;&lt; pos);
 }

 public static int clearNum(final int num, final int bitfield) {
 return bitfield ^ num;
 }

 }
</code></pre>

If there are some questions flying around, just write me an email.

Good Programming!

blocks|key|1397504|text|消除位移位及其复杂性，并对正确的and操作数使用LUT。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1397505|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Lookup_table^0|G|3|O|3|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]]|D|@$9|Q|A|R|1|S]]|E|$]]|$1|F|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]]]

Eliminate the bitshifting and its intricacies and use a <a href="http://en.wikipedia.org/wiki/Lookup_table" rel="nofollow noreferrer">LUT</a> for the right <code>and</code> operand.

What method would you use to determine if the the bit that represents 2^x is a 1 or 0 ?

Java: Checking if a bit is 0 or 1 in a long

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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您将使用什么方法来确定表示2^x的位是1还是0？

问Java:检查长整型中的位是0还是1
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回答 14

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问Java:检查长整型中的位是0还是1
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