我尝试将 double 转换为其二进制表示,但使用此 Long.toBinaryString(Double.doubleToRawLongBits(d))
并没有帮助,因为我有大量数字,Long 可以t 存储它们,即 2^900
.
I tried to convert a double to its binary representation, but using this Long.toBinaryString(Double.doubleToRawLongBits(d))
doesn't help, since I have large numbers, that Long can't store them i.e 2^900
.
Long.toBinaryString(Double.doubleToRawLongBits(d))
似乎工作得很好.
System.out.println("0: 0b" + Long.toBinaryString(Double.doubleToRawLongBits(0D)));
System.out.println("1: 0b" + Long.toBinaryString(Double.doubleToRawLongBits(1D)));
System.out.println("2: 0b" + Long.toBinaryString(Double.doubleToRawLongBits(2D)));
System.out.println("2^900: 0b" + Long.toBinaryString(Double.doubleToRawLongBits(Math.pow(2, 900))));
System.out.println("Double.MAX_VALUE: 0b" + Long.toBinaryString(Double.doubleToRawLongBits(Double.MAX_VALUE)));
/*
prints:
0: 0b0
1: 0b11111111110000000000000000000000000000000000000000000000000000
2: 0b100000000000000000000000000000000000000000000000000000000000000
2^900: 0b111100000110000000000000000000000000000000000000000000000000000
Double.MAX_VALUE: 0b111111111101111111111111111111111111111111111111111111111111111
*/
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