Jon*_*eet 410
假设你的意思是"内置":
int x = 100;
System.out.println(Integer.toBinaryString(x));
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请参阅整数文档.
(Long有一个类似的方法,BigInteger有一个实例方法,你可以指定基数.)
Moh*_*Ali 238
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你想要的程序格式.Integer.toString(int,representation);
Integer.toString(100,8) // prints 144 --octal representation
Integer.toString(100,2) // prints 1100100 --binary representation
Integer.toString(100,16) //prints 64 --Hex representation
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ada*_*shr 61
System.out.println(Integer.toBinaryString(343));
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M2X*_*M2X 21
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内容:
n = 5463
output = 0000 0000 0000 0000 0001 0101 0101 0111
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所以这就是我写的:
/**
* 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 % groupSize == 0)
result.append(" ");
}
result.replace(result.length() - 1, result.length(), "");
return result.toString();
}
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像这样调用它:
public static void main(String[] args) {
System.out.println(intToString(5463, 4));
}
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Sey*_*lal 10
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 & s)).replaceAll(" ", "0"));
System.out.println("b: " + String.format("%8s",
Integer.toBinaryString(0xFF & b)).replaceAll(" ", "0"));
}
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输出:
i: 00000000000000000000000000001101
s: 0000000000001101
b: 00001101
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老套:
int value = 28;
for(int i = 1, j = 0; i < 256; i = i << 1, j++)
System.out.println(j + " " + ((value & i) > 0 ? 1 : 0));
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小智 6
这是打印整数内部二进制表示的最简单方法。 例如:如果我们将 n 设为 17,那么输出将是:0000 0000 0000 0000 0000 0000 0001 0001
void bitPattern(int n) {
int mask = 1 << 31;
int count = 0;
while(mask != 0) {
if(count%4 == 0)
System.out.print(" ");
if((mask&n) == 0)
System.out.print("0");
else
System.out.print("1");
count++;
mask = mask >>> 1;
}
System.out.println();
}
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检查出这个逻辑可以将数字转换为任何基数
public static void toBase(int number, int base) {
String binary = "";
int temp = number/2+1;
for (int j = 0; j < temp ; j++) {
try {
binary += "" + number % base;
number /= base;
} catch (Exception e) {
}
}
for (int j = binary.length() - 1; j >= 0; j--) {
System.out.print(binary.charAt(j));
}
}
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要么
StringBuilder binary = new StringBuilder();
int n=15;
while (n>0) {
if((n&1)==1){
binary.append(1);
}else
binary.append(0);
n>>=1;
}
System.out.println(binary.reverse());
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