Question

对于原始值，Java的==运算符可以非对称，以便x == y，但是y != x，其中{{1} }和x都是某些（可能不同的）基元类型的变量？

修改

好的，我会说实话的原因。在Java Puzzlers book中，有一个谜题＃87涉及基本类型的y运算符。

它有三个部分，一个是挑战读者，以找到==运算符不反映的情况，第二个是查找==不可传递的情况。我找到了两者的解决方案，但我不知道如何解决第三个问题，其定义如下：

==

我没有这本书，所以我无法查找解决方案，我只查看了源文件，其中不包含解决方案。

Answer 1

没有。对于所有基元，==符号为commutative。也就是说，

（x == y）=＆gt; （y == x）

Answer 2

== 每个类型对称

来自规范：

如果操作数表达式没有边，则相等运算符是可交换的的效果。

Answer 3

对于相等和比较，只有一些特殊值。这些是Float.NaN，Double.NaN，-0.0f和-0.0。

这些特殊的原因是他们没有遵循一些正常的平等规则或者比较

public static void main(String... args) {
    printComparisons("Float.NaN", Float.NaN);
    printComparisons("Double.NaN", Double.NaN);
    printComparisons("-0.0f", -0.0f);
    printComparisons("-0.0", -0.0);
}

private static void printComparisons(String desc, float v) {
    System.out.println("[ " + desc + " ]");
    System.out.println(v + " == " + v + " is " + (v == v));
    System.out.println(v + " != " + v + " is " + (v != v));
    System.out.println(v + " == 0.0  is " + (v == 0.0));
    System.out.println(v + " < 0 is " + (v < 0));
    System.out.println(v + " > 0 is " + (v > 0));
    System.out.println("Float.compareTo(" + v + ", 0) is " + Float.compare(v, 0));
    System.out.println();
}

private static void printComparisons(String desc, double v) {
    System.out.println("[ " + desc + " ]");
    System.out.println(v + " == " + v + " is " + (v == v));
    System.out.println(v + " != " + v + " is " + (v != v));
    System.out.println(v + " == 0.0  is " + (v == 0.0));
    System.out.println(v + " < 0 is " + (v < 0));
    System.out.println(v + " > 0 is " + (v > 0));
    System.out.println("Double.compareTo(" + v + ", 0) is " + Double.compare(v, 0));
    System.out.println();
}

打印

[ Float.NaN ]
NaN == NaN is false
NaN != NaN is true
NaN == 0.0  is false
NaN < 0 is false
NaN > 0 is false
Float.compareTo(NaN, 0) is 1

[ Double.NaN ]
NaN == NaN is false
NaN != NaN is true
NaN == 0.0  is false
NaN < 0 is false
NaN > 0 is false
Double.compareTo(NaN, 0) is 1

[ -0.0f ]
-0.0 == -0.0 is true
-0.0 != -0.0 is false
-0.0 == 0.0  is true
-0.0 < 0 is false
-0.0 > 0 is false
Float.compareTo(-0.0, 0) is -1

[ -0.0 ]
-0.0 == -0.0 is true
-0.0 != -0.0 is false
-0.0 == 0.0  is true
-0.0 < 0 is false
-0.0 > 0 is false
Double.compareTo(-0.0, 0) is -1

总结。

NaN不是＆gt;，＆lt;，==，＆gt; =或＆lt; = 0.0但是compareTo(NaN,0)返回1
0.0 = = 0.0但是对于compareTo（-0.0,0.0），返回-1

Answer 4

我不知道任何基本类型都适用的情况。

Java的==运算符可以对原始值不对称吗？

4 个答案: