Switch Case Java 中的编译时间常数用法 [英] Compile Time Constant Usage in Switch Case Java

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本文介绍了Switch Case Java 中的编译时间常数用法的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

case的使用规则说:

  1. case 表达式的计算结果必须为 Compile Time Constant.

case(t) 表达式必须与 switch(t) 的类型相同,其中 t是类型(字符串).

case(t) expression must have same type as that of switch(t), where t is the type (String).

如果我运行这段代码:

public static void main(String[] args) {
    final String s=new String("abc");
    switch(s)
    {
       case (s):System.out.println("hi");
    }

}

编译错误为:"case expression must be a constant expression"另一方面,如果我尝试使用 final String s="abc";,它可以正常工作.

It gives Compile-error as: "case expression must be a constant expression" On the other hand if i try it with final String s="abc";, it works fine.

据我所知,String s=new String("abc") 是对位于堆上的 String 对象的引用.而 s 本身就是一个编译时常量.

As per my knowledge,String s=new String("abc") is a reference to a String object located on heap. And s itself is a compile-time constant.

是不是说final String s=new String("abc");不是编译时间常数?

Does it mean that final String s=new String("abc");isn't compile time constant?

推荐答案

用这个,

    String s= new String("abc");
    final String lm = "abc";

    switch(s)
    {
       case lm:
           case "abc": //This is more precise as per the comments
           System.out.println("hi");
           break;
    }

根据 文档

原始类型或 String 类型的变量,即 final 和用编译时常量表达式(§15.28)初始化,是称为常量变量

A variable of primitive type or type String, that is final and initialized with a compile-time constant expression (§15.28), is called a constant variable

问题是您的代码 final String s= new String("abc"); 没有初始化常量变量.

The problem is your code final String s= new String("abc"); does not initializes a constant variable.

这篇关于Switch Case Java 中的编译时间常数用法的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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