将Double转换为Numerics.BigInteger时的数字更改 [英] Number changes when converting Double to Numerics.BigInteger

问题描述

 使用System.Numerics; 
 
 double doubleNumber = Math.Pow（1000，99）; // = 1.0E + 297 
 BigInteger bigBase =（BigInteger）bigNumber; // = 1000000000000000017652801462756379714374878780719864776839443139119744823869255243069012222883470359078822072829219411228534934402712624705615450492327979456500795456339201761949451160807447294527656222743617592048849967890105831362861792425329827928397252374398383022243308510390698430058459037696 
  

<强>为什么是的BigInteger 值不仅仅是1卡吉（1,000，其次是一个袋子0）？

解决方案

BigInteger的实现使用大量的位移，所以当你将一个double转换成一个BigInteger时，双重表示有一个有限的位（双位为64位），所以并不是所有的位都准确地表示这个值。 p>

而不是使用Math.Pow，您应该使用

  var bigBase = BigInteger.Pow（1000，99）; 
  

另外，显式地将数字类型转换为BigInteger与使用BigInteger构造函数相同： p>

  var bigBase =（BigInteger）doubleNumber; 
  

相当于：

  var bigBase = new BigInteger（doubleNumber）; 
  

using System.Numerics;

double doubleNumber = Math.Pow(1000, 99); // = 1.0E+297
BigInteger bigBase = (BigInteger)bigNumber; // = 1000000000000000017652801462756379714374878780719864776839443139119744823869255243069012222883470359078822072829219411228534934402712624705615450492327979456500795456339201761949451160807447294527656222743617592048849967890105831362861792425329827928397252374398383022243308510390698430058459037696

Why is the BigInteger value not just 1 kagillion (1,000 followed by a bagillion 0's)?

解决方案

The implementation of BigInteger uses a lot of bit shifting, so when you convert a double to a BigInteger, the double representation has a finite set of bits (64 bits in for a double), so not all the bits are there to accurately represent the value.

Instead of using Math.Pow, you should be using

var bigBase = BigInteger.Pow(1000, 99);

Also, explicitly casting a numeric type to a BigInteger is the same as using the BigInteger constructor:

 var bigBase = (BigInteger)doubleNumber;

Is equivalent to:

var bigBase = new BigInteger(doubleNumber);

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