C#双精度问题

发布于 2024-09-18 01:11:42 字数 496 浏览 4 评论 0原文

想象一下 a - b c（a、b、c 是 C# 双精度数）。是否保证a < b + c？

谢谢！

编辑
假设不会发生算术溢出，与以下示例不同：

double a = 1L << 53;
double b = 1;
double c = a;

Console.WriteLine(a - b < c); // Prints True
Console.WriteLine(a < b + c); // Prints False

想象一下 Math.Abs(a) Math.Abs(a) 1.0&& Math.Abs(b) < 1.0&& Math.Abs(c) < 1.0

原文

Imagine that a - b < c (a, b, c are C# doubles). Is it guaranteed that a < b + c?

Thanks!

EDIT
Let's say that the arithmetical overflow doesn't occur unlike the following example:

double a = 1L << 53;
double b = 1;
double c = a;

Console.WriteLine(a - b < c); // Prints True
Console.WriteLine(a < b + c); // Prints False

Imagine that Math.Abs(a) < 1.0 && Math.Abs(b) < 1.0 && Math.Abs(c) < 1.0

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迟月 2024-09-25 01:11:42

不。假设a = c，一个非常大的数，而b 是一个非常小的数。 a - b 的表示形式可能小于 a，但 a + b 与 a 非常接近（甚至更大）它最终仍然可以最精确地表示为 a。

这是一个例子：

double a = 1L << 53;
double b = 1;
double c = a;

Console.WriteLine(a - b < c); // Prints True
Console.WriteLine(a < b + c); // Prints False

编辑：

这是另一个例子，与您编辑的问题相匹配：

double a = 1.0;
double b = 1.0 / (1L << 53);
double c = a;

Console.WriteLine(a - b < c); // Prints True
Console.WriteLine(a < b + c); // Prints False

换句话说，当我们从 1 中减去一个非常小的数字时，我们得到的结果小于 1。当我们将相同的数字加到 1 时，我们就得到1.由于双精度的限制返回。

No. Suppose a = c, a very large number, and b is a very small number. It's possible that a - b has a representation less than a, but a + b is so close to a (and bigger) that it still ends up being most precisely representable as a.

Here's an example:

double a = 1L << 53;
double b = 1;
double c = a;

Console.WriteLine(a - b < c); // Prints True
Console.WriteLine(a < b + c); // Prints False

EDIT:

Here's another example, which matches your edited question:

double a = 1.0;
double b = 1.0 / (1L << 53);
double c = a;

Console.WriteLine(a - b < c); // Prints True
Console.WriteLine(a < b + c); // Prints False

In other words, when we subtract a very small number from 1, we get a result less than 1. When we add the same number to 1, we just get 1 back due to the limitations of double precision.

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︶葆Ⅱㄣ 2024-09-25 01:11:42

不，并不总是：

        double a = double.MaxValue;
        double b = double.MaxValue;
        double c = 0.1;
        Console.WriteLine(a - b < c); // True
        Console.WriteLine(a < b + c); // False

no not always:

        double a = double.MaxValue;
        double b = double.MaxValue;
        double c = 0.1;
        Console.WriteLine(a - b < c); // True
        Console.WriteLine(a < b + c); // False

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