BigInteger如何存储 [英] How are BigInteger stored

问题描述

我需要生成512位BigInts，但是我不确定下面两个中的哪一个是正确的:

I need to generate 512 bit BigInts, but I'm not sure which of the two below is true:

512位表示1010101010...001010的512位数字，然后将其转换为它代表的小数?

512 bits means 512 digits of 1010101010...001010 which are then converted to the decimal it represents?

或者它表示0-9的512位数字，所以从根本上讲是一个0到9范围内的512位数字?类似于12414124124 .... 54543 = 512位数字.

Or does it mean 512 digits of 0-9, so basicly a 512-digit number with digits ranging from 0-9? Something like 12414124124....54543=512 digits.

推荐答案

从源代码开始，它们存储在int数组中

From sourcecode, they are stored in an int array

此BigInteger的大小，按大端顺序排列:此数组的第零个元素是该大小中最重要的int.幅度必须是最小"的，因为最重要的int(mag [0])必须不为零.必须确保每个BigInteger值都有一个唯一的表示形式.请注意，这意味着BigInteger零具有一个零长度的mag数组.

The magnitude of this BigInteger, in big-endian order: the zeroth element of this array is the most-significant int of the magnitude. The magnitude must be "minimal" in that the most-significant int (mag[0]) must be non-zero. This is necessary to ensure that there is exactly one representation for each BigInteger value. Note that this implies that the BigInteger zero has a zero-length mag array.

118 
119     int[] mag;
120 
121     // These "redundant fields" are initialized with recognizable nonsense
122     // values, and cached the first time they are needed (or never, if they
123     // aren't needed).
124 

这篇关于BigInteger如何存储的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！