Java 8:class.getName()和String文字之间的区别 [英] java 8: difference between class.getName() and String literal

问题描述

我正在处理开关盒.

如果我们使用class.getName()，则出现如下错误:大小写表达式必须是常量表达式":

If we use class.getName(), then, I am getting error that "case expressions must be constant expressions" as follows:

switch(param.getClass().getName())
    {
        case String.class.getName():
            // to do
            break;
    }

即使执行以下操作，也要在常量中使用字符串类名称，然后也会出现相同的错误:

Even if we do following, take string class name in a constant, then also, getting same error:

public static final String PARAM_NAME = String.class.getName();
switch(param.getClass().getName())
    {
        case PARAM_NAME:
            // to do
            break;
    }

但是，如果执行以下操作，请使用字符串文字"java.lang.String"，则不会出现错误:

But, if I do following, use the string literal "java.lang.String", there is not error:

public static final String PARAM_NAME = "java.lang.String";

任何人都可以解释一下，为什么它不受理前两个案件而不受理最后一个案件?预先感谢.

Can anybody please explain this, why its not taking first two cases and taking the last one? Thanks in advance.

推荐答案

classObject.getName()是方法调用，并且根据定义，方法调用的结果不是编译时常量.字符串文字是一个编译时常量.

classObject.getName() is a method call, and the results of method calls are by definition not compile-time constants. A string literal is a compile-time constant.

请注意，尽管在许多情况下都可以将static final引用作为程序生命周期的常量，但switch必须在编译时对其选项进行硬编码. case目标的值必须是枚举值或(编译时)

Note that while many situations could take a static final reference as a constant for the lifetime of the program, a switch has to have its options hard-coded at compile-time. The value of a case target must be either an enum value or a (compile-time) ConstantExpression.

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