blocks|key|2421494|text|按以下方式计算前导零的数目：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2421495|int+lz+=+8;
while+(i)
{
++++lz--;
++++i+>>>=+1;
}|code-block|syntax|javascript|2421496|当然，这假设数字不超过255，否则，您将得到负面的结果。|2421497|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|$]]

Count the number of leading zeros as follows:

<pre><code>int lz = 8;
while (i)
{
 lz--;
 i &gt;&gt;&gt;= 1;
}
</code></pre>

Of course, this supposes the number doesn't exceed 255, otherwise, you would get negative results.

blocks|key|4040963|text|怎么样|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4040964|int+lz+=+Integer.numberOfLeadingZeros(i+&+0xFF)+-+24;
int+tz+=+Integer.numberOfLeadingZeros(i+%7C+0x100);+//+max+is+8.|code-block|syntax|javascript|4040965|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|$]]

How about

<pre><code>int lz = Integer.numberOfLeadingZeros(i &amp; 0xFF) - 24;
int tz = Integer.numberOfLeadingZeros(i | 0x100); // max is 8.
</code></pre>

blocks|key|4017218|text|public+class+UtilsInt+{

++++int+leadingZerosInt(int+i)+{
++++++++return+leadingZeros(i,Integer.SIZE);
++++}

++++/**
+++++*+use+recursion+to+find+occurence+of+first+set+bit
+++++*+rotate+right+by+one+bit+&+adjust+complement
+++++*+check+if+rotate+value+is+not+zero+if+so+stop+counting/recursion
+++++*+@param+i+-+integer+to+check+
+++++*+@param+maxBitsCount+-+size+of+type+(in+this+case+int)
+++++*+++++++++++++++++++++++if+we+want+to+check+only+for:
+++++*+++++++++++++++++++++++++positive+values+we+can+set+this+to+Integer.SIZE+/+2+++
+++++*+++++++++++++++++++++++++(as+int+is+signed++in+java+-+positive+values+are+in+L16+bits)
+++++*/
+++++private+synchronized+int+leadingZeros(int+i,+int+maxBitsCount)+{
+++++++++try+{
++++++++++++logger.debug("checking+if+bit:+"%2B+maxBitsCount+
++++++++++++++++++++++++++++++++%2B+"+is+set+%7C+"+%2B+UtilsInt.intToString(i,8));
++++++++++++return+(i+>>>=+1)+!=+0+?+leadingZeros(i,+--maxBitsCount)+:+maxBitsCount;
+++++++++}+finally+{
+++++++++++++if(i==0)+logger.debug("bits+in+this+integer+from:+"+%2B+--maxBitsCount+
+++++++++++++++++++++++++++++++%2B+"+up+to+last+are+not+set+(i'm+counting+from+msb->lsb)");
+++++++++}
++++}
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4017219|测试声明：|unstyled|4017220|int+leadingZeros+=+new+UtilsInt.leadingZerosInt(255);+//+8|4017221|测试输出：|4017222|checking+if+bit:+32+is+set+%7C00000000+00000000+00000000+11111111
checking+if+bit:+31+is+set+%7C00000000+00000000+00000000+01111111
checking+if+bit:+30+is+set+%7C00000000+00000000+00000000+00111111
checking+if+bit:+29+is+set+%7C00000000+00000000+00000000+00011111
checking+if+bit:+28+is+set+%7C00000000+00000000+00000000+00001111
checking+if+bit:+27+is+set+%7C00000000+00000000+00000000+00000111
checking+if+bit:+26+is+set+%7C00000000+00000000+00000000+00000011
checking+if+bit:+25+is+set+%7C00000000+00000000+00000000+00000001
bits+in+this+integer+from:+24+up+to+last+are+not+set+(i'm+counting+from+msb->lsb)|4017223|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|P|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$B|C]]|$1|I|3|J|5|F|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$B|C]]|$1|M|3|-4|5|F|7|T|8|@]|9|@]|A|$]]]|N|$]]

<pre><code>public class UtilsInt {

 int leadingZerosInt(int i) {
 return leadingZeros(i,Integer.SIZE);
 }

 /**
 * use recursion to find occurence of first set bit
 * rotate right by one bit &amp; adjust complement
 * check if rotate value is not zero if so stop counting/recursion
 * @param i - integer to check 
 * @param maxBitsCount - size of type (in this case int)
 * if we want to check only for:
 * positive values we can set this to Integer.SIZE / 2 
 * (as int is signed in java - positive values are in L16 bits)
 */
 private synchronized int leadingZeros(int i, int maxBitsCount) {
 try {
 logger.debug("checking if bit: "+ maxBitsCount 
 + " is set | " + UtilsInt.intToString(i,8));
 return (i &gt;&gt;&gt;= 1) != 0 ? leadingZeros(i, --maxBitsCount) : maxBitsCount;
 } finally {
 if(i==0) logger.debug("bits in this integer from: " + --maxBitsCount 
 + " up to last are not set (i'm counting from msb-&gt;lsb)");
 }
 }
}
</code></pre>

test statement: 

<pre><code>int leadingZeros = new UtilsInt.leadingZerosInt(255); // 8
</code></pre>

test output: 

<pre><code>checking if bit: 32 is set |00000000 00000000 00000000 11111111
checking if bit: 31 is set |00000000 00000000 00000000 01111111
checking if bit: 30 is set |00000000 00000000 00000000 00111111
checking if bit: 29 is set |00000000 00000000 00000000 00011111
checking if bit: 28 is set |00000000 00000000 00000000 00001111
checking if bit: 27 is set |00000000 00000000 00000000 00000111
checking if bit: 26 is set |00000000 00000000 00000000 00000011
checking if bit: 25 is set |00000000 00000000 00000000 00000001
bits in this integer from: 24 up to last are not set (i'm counting from msb-&gt;lsb)
</code></pre>

I need <code>int 32</code> in binary as <code>00100000</code> or <code>int 127</code> in binary <code>0111 1111</code>.
The variant <code>Integer.toBinaryString</code> returns results only from 1.
If I build the for loop this way:

<pre><code>for (int i= 32; i &lt;= 127; i + +) {
System.out.println (i); 
System.out.println (Integer.toBinaryString (i));
}
</code></pre>

And from binary numbers I need the number of leading zeros (count leading zeros (clz) or number of leading zeros (nlz)) I really meant the exact number of 0, such ex: at 00100000 -> 2 and at 0111 1111 - > 1

count leading zeros (clz) or number of leading zeros (nlz) in Java

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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我需要二进制的int 32，就像二进制0111 1111中的00100000或int 127。变体Integer.toBinaryString只返回1的结果。如果以这种方式构建for循环：for (int i= 32; i <= 127; i + +) {System.out.println (i); System.o...

问在Java中计数前导零(clz)或前导零数(nlz)
EN

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问在Java中计数前导零(clz)或前导零数(nlz)EN