import java.math.BigInteger; public class TestBigInteger { public static void main(String[] args) { //初始 BigInteger num = new BigInteger("0"); num = num.setBit(2); num = num.setBit(1); System.out.println(num); System.out.println(num.testBit(2)); System.out.println(num.testBit(1)); System.out.println(num.testBit(3)); } }
返回的结果是:
6
true
true
false
为什么是6呢? 6= 2^2 + 2^1 其实计算的值是2的权的和
import java.math.*; public class BigIntegerDemo { public static void main(String[] args) { // create a BigInteger object BigInteger bi; // create 2 boolean objects Boolean b1, b2; bi = new BigInteger("10"); // perform testbit on bi at index 2 and 3 b1 = bi.testBit(2); b2 = bi.testBit(3); String str1 = "Test Bit on " + bi + " at index 2 returns " +b1; String str2 = "Test Bit on " + bi + " at index 3 returns " +b2; // print b1, b2 values System.out.println( str1 ); System.out.println( str2 ); } }
让我们编译和运行上面的程序,这将产生以下结果: Test Bit on 10 at index 2 returns false Test Bit on 10 at index 3 returns truejava.math.BigInteger.testBit(int n) 当且仅当所指定的位被置位时返回true。它计算方式为 (this & (1<<n)) != 0).
总结一下:
& 是比较位的, 两个都位1 , 则保留,
1<<2 = 4 二进制(100) 100 & 1010 = 0
1<<3 =8 二进制(1000) 1000& 1010 = 8