为什么我们不能使用 '==' 来比较两个浮点数或双精度数

Compares two Float objects numerically. There are two ways in which comparisons performed by this method differ from those performed by the Java language numerical comparison operators (<, <=, ==, >= >) when applied to primitive float values:

• 此方法认为 Float.NaN 等于其自身并且大于所有其他浮点值(包括 Float.POSITIVE_INFINITY).
• 此方法认为0.0f大于-0.0f.

这确保了此方法强加的 Float 对象的自然顺序与 equals 一致.

考虑以下代码:

    System.out.println(-0.0f == 0.0f); //true
    System.out.println(Float.compare(-0.0f, 0.0f) == 0 ? true : false); //false      
    System.out.println(Float.NaN == Float.NaN);//false
    System.out.println(Float.compare(Float.NaN, Float.NaN) == 0 ? true : false); //true
    System.out.println(-0.0d == 0.0d); //true
    System.out.println(Double.compare(-0.0d, 0.0d) == 0 ? true : false);//false     
    System.out.println(Double.NaN == Double.NaN);//false
    System.out.println(Double.compare(Double.NaN, Double.NaN) == 0 ? true : false);//true        

输出不正确，因为不是数字的东西，根本就不是数字，从数字比较的角度来看，应该被视为相等.0=-0.

public static int compare(float f1, float f2) {
   if (f1 < f2)
        return -1;           // Neither val is NaN, thisVal is smaller
    if (f1 > f2)
        return 1;            // Neither val is NaN, thisVal is larger

    int thisBits = Float.floatToIntBits(f1);
    int anotherBits = Float.floatToIntBits(f2);

    return (thisBits == anotherBits ?  0 : // Values are equal
            (thisBits < anotherBits ? -1 : // (-0.0, 0.0) or (!NaN, NaN)
             1));                          // (0.0, -0.0) or (NaN, !NaN)
}