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[Java教程]Java你可能不知道的事系列(1)


概述

本类文章会不段更新分析学习到的经典面试题目,在此记录下来便于自己理解。如果有不对的地方还请各位观众拍砖。 
今天主要分享一下常用的字符串的几个题目,相信学习java的小伙伴们对String类是再熟悉不过了,今天我们就来和她再次邂逅,好了下面开始。

先来说说String特点

String是不可变的常量,每当我们创建一个字符串对象的时候,如果堆区的常量池里不存在这个字符串,就会创建一个存储在常量池里(String存的地方叫String pool),如果存在了,就直接把变量的地址指向常量池里,比如:String b = “abc” 这句话 内存表示如下。下面开始上题 
这里写图片描述

1.1

String s1 = new String("abc");String s2 = new String("abc"); System.out.println(s1 == s2);

 

输出结果是什么呢? 
从上面的图也大概说了jvm里面有堆、栈区。堆区里面主要存放的是局部变量,栈区里存放的是new出来的对象。==对于对象类型比较的是地址。所以在s1和s1是分别引用了堆里面new出来的不同对象的地址,图形理解如下

这里写图片描述

答案很明显了,地址不同 输出false.

1.2

String s1 = "abc";StringBuffer s2 = new StringBuffer(s1); System.out.println(s1.equals(s2));

这是true 还是false呢?答案是false。

首先s1变量引用了字符串”abc”,然后StringBuffer s2 = new StringBuffer(s1),新建了一个StringBuffer对象调用append()方法返回自身。调用String的equals方法。重点就是这个equals方法里有个instance of,必需是同一类型的才进行比较否则直接返回false。 
来看一下源码:

/**   * Compares this string to the specified object. The result is {@code   * true} if and only if the argument is not {@code null} and is a {@code   * String} object that represents the same sequence of characters as this   * object.   *   * @param anObject   *     The object to compare this {@code String} against   *   * @return {@code true} if the given object represents a {@code String}   *     equivalent to this string, {@code false} otherwise   *   * @see #compareTo(String)   * @see #equalsIgnoreCase(String)   */  public boolean equals(Object anObject) {    if (this == anObject) {      return true;    }    //关键点就在这里了    if (anObject instanceof String) {      String anotherString = (String) anObject;      int n = value.length;      if (n == anotherString.value.length) {        char v1[] = value;        char v2[] = anotherString.value;        int i = 0;        while (n-- != 0) {          if (v1[i] != v2[i])              return false;          i++;        }        return true;      }    }    return false;  }

 

1.3

下面的代码在内存会产生几个对象呢? 

String s1 = new String(“abc”); String s2 = new String(“abc”);

 

答案:3个 
有了上面的分析,相信大家都明白了,new了两个对象,加上string pool里的一个”abc”。

1.4

下面的代码输出结果是啥?

String s1 = "abc";String s2 = new String("abc");s2.intern();System.out.println(s1 ==s2);

 

我们可能对intern()这个方法不太熟悉,先来看看注释:

/**   * Returns a canonical representation for the string object.   * <p>   * A pool of strings, initially empty, is maintained privately by the   * class <code>String</code>.   * <p>   * When the intern method is invoked, if the pool already contains a   * string equal to this <code>String</code> object as determined by   * the {@link #equals(Object)} method, then the string from the pool is   * returned. Otherwise, this <code>String</code> object is added to the   * pool and a reference to this <code>String</code> object is returned.   * <p>   * It follows that for any two strings <code>s</code> and <code>t</code>,   * <code>s.intern()&nbsp;==&nbsp;t.intern()</code> is <code>true</code>   * if and only if <code>s.equals(t)</code> is <code>true</code>.   * <p>   * All literal strings and string-valued constant expressions are   * interned. String literals are defined in section 3.10.5 of the   * <cite>The Java&trade; Language Specification</cite>.   *   * @return a string that has the same contents as this string, but is   *     guaranteed to be from a pool of unique strings.   */  public native String intern();

 

注释好多我草,关键的是这句:

When the intern method is invoked, if the pool already contains a string equal to this String object as determined by the {@link #equals(Object)} method, then the string from the pool is returned. Otherwise, this String object is added to the pool and a reference to this String object is returned. 


大致就是说,如果常量池里不存在这个字符串,就创建一个并且返回地址,否则的话直接返回地址。

上面的代码第二行String s2 = new String(“abc”); s2其实是引用到了new的对象,虽然在第三行调用了intern方法,但是没有赋值给s2,所以s2的引用还是没有变。所以返回false。 
如果第三行代码改成s2 = s2.intern()就会返回true了。

 String s1 = "abc"; String s2 = new String("abc"); s2 = s2.intern(); System.out.println(s1==s2);

 

好了,今天就到这里。之后会继续分析。如果喜欢我的文章欢迎关注我。求赞,有问题可以评论,各位的支持是我最大的动力!!