blocks|key|757507|text|假设你指的是“内置”：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|757508|int+x+=+100;
System.out.println(Integer.toBinaryString(x));|code-block|syntax|javascript|757509|参见Integer+documentation。|offset|length|757510|(Long有一个类似的方法，BigInteger有一个实例方法，可以在其中指定基数。)|style|CODE|757511|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/8/docs/api/java/lang/Integer.html#toBinaryString-int-^0|0|0|2|L|0|0|1|4|E|A|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@$I|Z|J|10|1|11]]|A|$]]|$1|K|3|L|5|6|7|12|8|@$I|13|J|14|M|N]|$I|15|J|16|M|N]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]]]

Assuming you mean "built-in":

<pre><code>int x = 100;
System.out.println(Integer.toBinaryString(x));
</code></pre>

See <a href="http://docs.oracle.com/javase/8/docs/api/java/lang/Integer.html#toBinaryString-int-" rel="noreferrer">Integer documentation</a>.

(<code>Long</code> has a similar method, <code>BigInteger</code> has an instance method where you can specify the radix.)

blocks|key|757585|text|这里不需要只依赖于二进制或任何其他格式...一个灵活的内置函数可用于打印程序中所需的任何格式。Integer.toString(int,+representation)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|757586|Integer.toString(100,8)+//+prints+144+--octal+representation

Integer.toString(100,2)+//+prints+1100100+--binary+representation

Integer.toString(100,16)+//prints+64+--Hex+representation|code-block|syntax|javascript|757587|entityMap^0|1B|11|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|$]]

Here no need to depend only on binary or any other format... one flexible built in function is available That prints whichever format you want in your program.. <code>Integer.toString(int, representation)</code>
<pre><code>Integer.toString(100,8) // prints 144 --octal representation

Integer.toString(100,2) // prints 1100100 --binary representation

Integer.toString(100,16) //prints 64 --Hex representation
</code></pre>

blocks|key|757540|text|System.out.println(Integer.toBinaryString(343));|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|757541|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>System.out.println(Integer.toBinaryString(343));
</code></pre>

blocks|key|973195|text|我需要一些东西来很好地打印出来，并将每个n位的位分开。换句话说，显示前导零并显示如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|973196|n+=+5463
output+=+0000+0000+0000+0000+0001+0101+0101+0111|code-block|syntax|javascript|973197|下面是我写的内容：|973198|/**
+*+Converts+an+integer+to+a+32-bit+binary+string
+*+@param+number
+*++++++The+number+to+convert
+*+@param+groupSize
+*++++++The+number+of+bits+in+a+group
+*+@return
+*++++++The+32-bit+long+bit+string
+*/
public+static+String+intToString(int+number,+int+groupSize)+{
++++StringBuilder+result+=+new+StringBuilder();

++++for(int+i+=+31;+i+>=+0+;+i--)+{
++++++++int+mask+=+1+<<+i;
++++++++result.append((number+&+mask)+!=+0+?+"1"+:+"0");

++++++++if+(i+%25+groupSize+==+0)
++++++++++++result.append("+");
++++}
++++result.replace(result.length()+-+1,+result.length(),+"");

++++return+result.toString();
}|973199|像这样调用它：|973200|public+static+void+main(String[]+args)+{
++++System.out.println(intToString(5463,+4));
}|973201|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

I needed something to print things out nicely and separate the bits every n-bit. In other words display the leading zeros and show something like this:

<pre><code>n = 5463
output = 0000 0000 0000 0000 0001 0101 0101 0111
</code></pre>

So here's what I wrote:

<pre><code>/**
 * Converts an integer to a 32-bit binary string
 * @param number
 * The number to convert
 * @param groupSize
 * The number of bits in a group
 * @return
 * The 32-bit long bit string
 */
public static String intToString(int number, int groupSize) {
 StringBuilder result = new StringBuilder();

 for(int i = 31; i &gt;= 0 ; i--) {
 int mask = 1 &lt;&lt; i;
 result.append((number &amp; mask) != 0 ? "1" : "0");

 if (i % groupSize == 0)
 result.append(" ");
 }
 result.replace(result.length() - 1, result.length(), "");

 return result.toString();
}
</code></pre>

Invoke it like this:

<pre><code>public static void main(String[] args) {
 System.out.println(intToString(5463, 4));
}
</code></pre>

blocks|key|757835|text|public+static+void+main(String[]+args)+
{
++++int+i+=+13;
++++short+s+=+13;
++++byte+b+=+13;

++++System.out.println("i:+"+%2B+String.format("%2532s",+
++++++++++++Integer.toBinaryString(i)).replaceAll("+",+"0"));

++++System.out.println("s:+"+%2B+String.format("%2516s",+
++++++++++++Integer.toBinaryString(0xFFFF+&+s)).replaceAll("+",+"0"));

++++System.out.println("b:+"+%2B+String.format("%258s",+
++++++++++++Integer.toBinaryString(0xFF+&+b)).replaceAll("+",+"0"));

}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|757836|输出：|unstyled|757837|i:+00000000000000000000000000001101
s:+0000000000001101
b:+00001101|757838|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>public static void main(String[] args) 
{
 int i = 13;
 short s = 13;
 byte b = 13;

 System.out.println("i: " + String.format("%32s", 
 Integer.toBinaryString(i)).replaceAll(" ", "0"));

 System.out.println("s: " + String.format("%16s", 
 Integer.toBinaryString(0xFFFF &amp; s)).replaceAll(" ", "0"));

 System.out.println("b: " + String.format("%8s", 
 Integer.toBinaryString(0xFF &amp; b)).replaceAll(" ", "0"));

}
</code></pre>

Output:

<pre><code>i: 00000000000000000000000000001101
s: 0000000000001101
b: 00001101
</code></pre>

blocks|key|973488|text|简单且相当容易的解决方案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|973489|public+static+String+intToBinaryString(int+integer,+int+numberOfBits)+{

++++if+(numberOfBits+>+0)+{+++++//+To+prevent+FormatFlagsConversionMismatchException.

++++++++String+nBits+=+String.format("%25"+%2B+numberOfBits+%2B+"s",++++++//+Int+to+bits+conversion
++++++++++++++++Integer.toBinaryString(integer))
++++++++++++++++.replaceAll("+","0");+

++++++++return+nBits;+++//+returning+the+Bits+for+the+given+int.
++++}

++++return+null;++++++++//+if+the+numberOfBits+is+not+greater+than+0,+returning+null.
}|code-block|syntax|javascript|973490|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|$]]

Simple and pretty easiest solution.

<pre><code>public static String intToBinaryString(int integer, int numberOfBits) {

 if (numberOfBits &gt; 0) { // To prevent FormatFlagsConversionMismatchException.

 String nBits = String.format("%" + numberOfBits + "s", // Int to bits conversion
 Integer.toBinaryString(integer))
 .replaceAll(" ","0"); 

 return nBits; // returning the Bits for the given int.
 }

 return null; // if the numberOfBits is not greater than 0, returning null.
}
</code></pre>

blocks|key|973434|text|使用32位显示掩模的解决方案，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|973435|public+static+String+toBinaryString(int+n){

++++StringBuilder+res=new+StringBuilder();
++++//res=+Integer.toBinaryString(n);+or
++++int+displayMask=1<<31;
++++for+(int+i=1;i<=32;i%2B%2B){
++++++++res.append((n+&+displayMask)==0?'0':'1');
++++++++n=n<<1;
++++++++if+(i%258==0)+res.append('+');
++++}

++++return+res.toString();
}


+System.out.println(BitUtil.toBinaryString(30));


O/P:
00000000+00000000+00000000+00011110+|code-block|syntax|javascript|973436|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|$]]

Solution using 32 bit display mask,

<pre><code>public static String toBinaryString(int n){

 StringBuilder res=new StringBuilder();
 //res= Integer.toBinaryString(n); or
 int displayMask=1&lt;&lt;31;
 for (int i=1;i&lt;=32;i++){
 res.append((n &amp; displayMask)==0?'0':'1');
 n=n&lt;&lt;1;
 if (i%8==0) res.append(' ');
 }

 return res.toString();
}


 System.out.println(BitUtil.toBinaryString(30));


O/P:
00000000 00000000 00000000 00011110 
</code></pre>

blocks|key|973319|text|给定int+x的二进制表示，带有左填充的零：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|973320|org.apache.commons.lang3.StringUtils.leftPad(Integer.toBinaryString(x),+32,+'0')|code-block|syntax|javascript|973321|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|$]]

Binary representation of given int x with left padded zeros:

<pre><code>org.apache.commons.lang3.StringUtils.leftPad(Integer.toBinaryString(x), 32, '0')
</code></pre>

blocks|key|757703|text|++++for(int+i+=+1;+i+<=+256;+i%2B%2B)
++++{
++++++++System.out.print(i+%2B+"+");+//show+integer
++++++++System.out.println(Integer.toBinaryString(i)+%2B+"+");+//show+binary
++++++++System.out.print(Integer.toOctalString(i)+%2B+"+");+//show+octal
++++++++System.out.print(Integer.toHexString(i)+%2B+"+");+//show+hex

++++}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|757704|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code> for(int i = 1; i &lt;= 256; i++)
 {
 System.out.print(i + " "); //show integer
 System.out.println(Integer.toBinaryString(i) + " "); //show binary
 System.out.print(Integer.toOctalString(i) + " "); //show octal
 System.out.print(Integer.toHexString(i) + " "); //show hex

 }
</code></pre>

blocks|key|757716|text|尝试这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|757717|public+class+Bin+{
++public+static+void+main(String[]+args)+{
++++System.out.println(toBinary(0x94,+8));
++}

++public+static+String+toBinary(int+a,+int+bits)+{
++++if+(--bits+>+0)
++++++++return+toBinary(a>>1,+bits)%2B((a&0x1)==0?"0":"1");
++++else+
++++++++return+(a&0x1)==0?"0":"1";
++}|code-block|syntax|javascript|757718|}|757719|10010100|757720|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|$]]

Try this way:

<pre><code>public class Bin {
 public static void main(String[] args) {
 System.out.println(toBinary(0x94, 8));
 }

 public static String toBinary(int a, int bits) {
 if (--bits &gt; 0)
 return toBinary(a&gt;&gt;1, bits)+((a&amp;0x1)==0?"0":"1");
 else 
 return (a&amp;0x1)==0?"0":"1";
 }
</code></pre>

}

10010100

blocks|key|973390|text|输入任意十进制数作为输入。之后，我们进行模运算和除法运算，将给定的输入转换为二进制数。以下是将Integer值转换为二进制的Java程序的源代码，以及该二进制的十进制数的位数。该Java程序已成功编译并在Windows系统上运行。程序输出也如下所示。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|973391|++++public+static+void+main(String[]+args)+{
++++++++Scanner+sc+=+new+Scanner(System.in);
++++++++int+integer+;
++++++++String+binary+=+"";++++//+++here+we+count+""+or+null+
+++++++++++++++++++++++++++++++//+++just+String+binary+=+null;
++++++++System.out.print("Enter+the+binary+Number:+");
++++++++integer+=+sc.nextInt();

++++++++while(integer>0)
++++++++{
++++++++++++int+x+=+integer+%25+2;
++++++++++++binary+=+x+%2B+binary;
++++++++++++integer+=+integer+/+2;++
++++++++}
++++++++System.out.println("Your+binary+number+is+:+"%2Bbinary);
++++++++System.out.println("your+binary+length+:+"+%2B+binary.length());
++++}
}|code-block|syntax|javascript|973392|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|$]]

Enter any decimal number as an input. After that we operations like modulo and division to convert the given input into binary number.
Here is the source code of the Java Program to Convert Integer Values into Binary and the bits number of this binary for his decimal number.
The Java program is successfully compiled and run on a Windows system. The program output is also shown below.

<pre><code> public static void main(String[] args) {
 Scanner sc = new Scanner(System.in);
 int integer ;
 String binary = ""; // here we count "" or null 
 // just String binary = null;
 System.out.print("Enter the binary Number: ");
 integer = sc.nextInt();

 while(integer&gt;0)
 {
 int x = integer % 2;
 binary = x + binary;
 integer = integer / 2; 
 }
 System.out.println("Your binary number is : "+binary);
 System.out.println("your binary length : " + binary.length());
 }
}
</code></pre>

blocks|key|973527|text|由于没有接受任何答案，也许您的问题是关于如何在二进制文件中存储整数。你要找的可能就是java.io.DataOutputStream：https://docs.oracle.com/javase/8/docs/api/java/io/DataOutputStream.html|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|973528|DataOutputStream+os+=+new+DataOutputStream(outputStream);
os.writeInt(42);
os.flush();
os.close();|code-block|syntax|javascript|973529|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.oracle.com/javase/8/docs/api/java/io/DataOutputStream.html^0|1V|1Z|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

Since no answer is accepted, maybe your question was about how to store an integer in an binary-file.
java.io.DataOutputStream might be what you're looking for: <a href="https://docs.oracle.com/javase/8/docs/api/java/io/DataOutputStream.html" rel="nofollow noreferrer">https://docs.oracle.com/javase/8/docs/api/java/io/DataOutputStream.html</a>

<pre><code>DataOutputStream os = new DataOutputStream(outputStream);
os.writeInt(42);
os.flush();
os.close();
</code></pre>

blocks|key|758014|text|你可以使用位掩码(1<<+k)和do与数字运算！1+<<+k在k位置有一个位！|type|unstyled|depth|inlineStyleRanges|entityRanges|data|758015|private+void+printBits(int+x)+{
++++for(int+i+=+31;+i+>=+0;+i--)+{
++++++++if((x+&+(1+<<+i))+!=+0){
++++++++++++System.out.print(1);
++++++++}else+{
++++++++++++System.out.print(0);
++++++++}
++++}
++++System.out.println();
}|code-block|syntax|javascript|758016|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|$]]

You can use bit mask (1&lt;&lt; k) and do AND operation with number!
1 &lt;&lt; k has one bit at k position!

<pre><code>private void printBits(int x) {
 for(int i = 31; i &gt;= 0; i--) {
 if((x &amp; (1 &lt;&lt; i)) != 0){
 System.out.print(1);
 }else {
 System.out.print(0);
 }
 }
 System.out.println();
}
</code></pre>

blocks|key|758052|text|Integer.toString(+value，numbersystem)+-要使用的语法和传递的值|type|unstyled|depth|inlineStyleRanges|entityRanges|data|758053|Integer.toString(100,8)+//打印144+--八进制|758054|Integer.toString(100,2)+//打印1100100+--二进制|758055|Integer.toString(100,16)+//打印64+--十六进制|758056|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|L|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|M|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

Integer.toString(value,numbersystem) --- syntax to be used
and pass value
Integer.toString(100,8) // prints 144 --octal
Integer.toString(100,2) // prints 1100100 --binary
Integer.toString(100,16) //prints 64 --Hex

blocks|key|758100|text|这是我格式化Integer.toBinaryString方法输出的方式：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|758101|public+String+toBinaryString(int+number,+int+groupSize)+{
++++String+binary+=+Integer.toBinaryString(number);

++++StringBuilder+result+=+new+StringBuilder(binary);
++++for+(int+i+=+1;+i+<+binary.length();+i%2B%2B)+{
++++++++if+(i+%25+groupSize+==+0)+{
++++++++++++result.insert(binary.length()+-+i,+"+");
++++++++}
++++}
++++return+result.toString();
}|code-block|syntax|javascript|758102|toBinaryString(0xABFABF,+8)的结果是"10101011+11111010+10111111"|758103|用于toBinaryString(0xABFABF,+4)的是"1010+1011+1111+1010+1011+1111"|758104|entityMap^0|6|M|0|0|0|R|V|S|0|2|R|V|V|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|$]]

This is my way to format an output of the <code>Integer.toBinaryString</code> method:
<pre><code>public String toBinaryString(int number, int groupSize) {
 String binary = Integer.toBinaryString(number);

 StringBuilder result = new StringBuilder(binary);
 for (int i = 1; i &lt; binary.length(); i++) {
 if (i % groupSize == 0) {
 result.insert(binary.length() - i, &quot; &quot;);
 }
 }
 return result.toString();
}
</code></pre>
The result for the <code>toBinaryString(0xABFABF, 8)</code> is <code>&quot;10101011 11111010 10111111&quot;</code>
 
and for the <code>toBinaryString(0xABFABF, 4)</code> is <code>&quot;1010 1011 1111 1010 1011 1111&quot;</code>

blocks|key|973247|text|它使用强大的位操作来处理有符号和无符号的值，并在左边生成前几个零。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|973248|public+static+String+representDigits(int+num)+{

++++++++int+checkBit+=+1+<<+(Integer.SIZE+*+8+-+2+);++++//+avoid+the+first+digit++++++++
++++++++StringBuffer+sb+=+new+StringBuffer();

++++++++if+(num+<+0+)+{+++++//+checking+the+first+digit
++++++++++++sb.append("1");
++++++++}+else+{
++++++++++++sb.append("0");
++++++++}

++++++++while(checkBit+!=+0)+{++++++++++
++++++++++++if+((num+&+checkBit)+==+checkBit){
++++++++++++++++sb.append("1");
++++++++++++}+else+{
++++++++++++++++sb.append("0");
++++++++++++}+++++++++++
++++++++++++checkBit+>>=+1;+++++
++++++++}+++++++

++++++++return+sb.toString();
++++}|code-block|syntax|javascript|973249|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|$]]

It works with signed and unsigned values used powerful bit manipulation and generates the first zeroes on the left.

<pre><code>public static String representDigits(int num) {

 int checkBit = 1 &lt;&lt; (Integer.SIZE * 8 - 2 ); // avoid the first digit 
 StringBuffer sb = new StringBuffer();

 if (num &lt; 0 ) { // checking the first digit
 sb.append("1");
 } else {
 sb.append("0");
 }

 while(checkBit != 0) { 
 if ((num &amp; checkBit) == checkBit){
 sb.append("1");
 } else {
 sb.append("0");
 } 
 checkBit &gt;&gt;= 1; 
 } 

 return sb.toString();
 }
</code></pre>

I have a number and I want to print it in binary. I don't want to do it by writing an algorithm.
Is there any built-in function for that in Java?

Print an integer in binary format in Java

Windows

Java

我有一个数字，我想把它打印成二进制。我不想通过编写算法来做到这一点。Java中有没有内置的函数来实现这个功能？

问在Java中以二进制格式打印整数
EN

回答 16

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问在Java中以二进制格式打印整数EN

回答 16

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问在Java中以二进制格式打印整数
EN