我很感谢您的评论和参考，但我仍然不了解如何存储双精度型。我确实了解浮点数的工作原理，并认为double的效果相同，但是使用了另一个字节来提高精度。简而言之，这对我来说毫无意义：
double a = 5.01 is stored as 5.01 how is it able to do that?
since .1 can't be represented in binary?
此问题不是所引用问题的重复。在第一种情况下，.01存储为.0099999999787，在第二种情况下，.01存储为.01000000000000
println有时会绕一圈的评论似乎是合理的，但计算值似乎确实是.0100000000000
我有一个我不理解的以下示例：
double a = 6;
double b = 5.99;
double c = .99;
System.out.println((a-b) + " " + (a-c));
why is a - b = .0099999999999787 (which seems correct)
but a - c = 5.01 (which seems incorrect)
shouldn't a - c = 5.0099999999787 ?