C# 用 .Equals() 比较两个双精度

问题描述

我使用 ReShaper，当我用 == 比较两个双精度值时，它建议我应该使用 Math.具有公差的 ABS 方法.请参阅:https://www.jetbrains.com/help/resharper/2016.2/CompareOfFloatsByEqualityOperator.html

I use ReShaper and when I compare two double values with ==, it suggests that I should use the Math. ABS method with a tolerance. See: https://www.jetbrains.com/help/resharper/2016.2/CompareOfFloatsByEqualityOperator.html

这个例子

double d = 0.0;
double d2 = 0.0;
if (d == d2)
{
    /* some code */
}

然后转换为

double d = 0.0;
double d2 = 0.0;
if (Math.Abs(d - d2) < TOLERANCE)
{
    /* some code */
}

但我认为，对于开发人员来说，考虑正确的容差确实很复杂.所以我认为这可能在 Double.Equals() 方法中实现.

But I think it's really complicated for a developer to think about the right tolerance. So I thought this may be implemented in the Double.Equals() method.

但是这个方法是这样实现的

But this method is implemented like so

public override bool Equals(Object obj) {
    if (!(obj is Double)) {
        return false;
    }
    double temp = ((Double)obj).m_value;
    // This code below is written this way for performance reasons i.e the != and == check is intentional.
    if (temp == m_value) {
        return true;
    }
    return IsNaN(temp) && IsNaN(m_value);
}

public bool Equals(Double obj)
{
    if (obj == m_value) {
        return true;
    }
    return IsNaN(obj) && IsNaN(m_value);
}

这是为什么呢?比较双精度值的正确方法是什么?

推荐答案

你可以创建一个扩展方法

You could create an extension method

public static class DoubleExtension
{
    public static bool AlmostEqualTo(this double value1, double value2)
    {
        return Math.Abs(value1 - value2) < 0.0000001;
    }
}

并像这样使用它

doubleValue.AlmostEqualTo(doubleValue2)

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