了解 C# 和 Java 中的 IEEE-754 64 位定点表示

考虑以下 Java 代码：

public class Program {
    public static void main(String args[]) {
      double number = Double.MAX_VALUE;
      String formattedNumber = String.format("%f", number);
      System.out.println(formattedNumber);
    }
}

17976931348623157000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000000000000000000.000000

考虑等效的 C# 代码：

public class Program
{
    public static void Main(string[] args)
    {
        double value = double.MaxValue;
        Console.WriteLine(value.ToString("F"));
    }
}

1797693134862315708145274237317043567980705675258449965989174768031572607800285387605895586327668781715404589535143824642 343213268894641827684675467035375169860499105765512820762454900903893289440758685084551339423045832369032229481658085593321 23348274797826204144723168738177180919299881250404026184124858368.000

鉴于最大值Double是1.7976931348623157E+308，据我所知，Java输出是正确的；即浮点值实际上表示一个整数，其中前 17 位数字为 17976931348623157，后跟 292 个零。

注：转换double to BigInteger在 C# 中会产生相同的结果：

BigInteger value = (BigInteger)double.MaxValue;
Console.WriteLine(value);

1797693134862315708145274237317043567980705675258449965989174768031572607800285387605895586327668781715404589535143824642 343213268894641827684675467035375169860499105765512820762454900903893289440758685084551339423045832369032229481658085593321 23348274797826204144723168738177180919299881250404026184124858368

问题

• 为什么这些值差异很大，哪一个应该被认为是正确的？
• 如果 C# 实际上不正确，我如何获得与 Java 生成的正确或相同的输出？

鉴于 Double 的最大值为 1.7976931348623157E+308...

This is incorrect. The format used for Double is the binary64 or “double precision” format specified in the IEEE 754-2019 standard. In this format, a finite number is represented as ±f•2^e, where f, the fraction portion of the representation, is the number represented by a binary numeral with one digit (0 or 1) before the radix point and 52 bits after it and e, the exponent, is an integer in [−1022, 1023]. So the maximum representable finite value is +1.1111111111111111111111111111111111111111111111111111₂•2¹⁰²³, which equals +(2¹−2⁻⁵²)•2¹⁰²³ = 2¹⁰²⁴−2⁹⁷¹, which is exactly 179,769,313,486,231,570,814,527,423,731,704,356,798,070,567,525,844,996,598,917,476,803,157,260,780,028,538,760,589,558,632,766,878,171,540,458,953,514,382,464,234,321,326,889,464,182,768,467,546,703,537,516,986,049,910,576,551,282,076,245,490,090,389,328,944,075,868,508,455,133,942,304,583,236,903,222,948,165,808,559,332,123,348,274,797,826,204,144,723,168,738,177,180,919,299,881,250,404,026,184,124,858,368, so the C# output is correct.

Java 输出省略有效数字并隐藏真实值。这是因为 Java 规范规定对于默认格式“必须至少有一位数字来表示小数部分，除此之外，还有尽可能多的数字，以唯一区分参数值与类型的相邻值double.” https://docs.oracle.com/en/java/javase/17/docs/api/java.base/java/lang/Double.html#toString(double)