Detailed explanation of strings in Java
String
In java, strings are treated as objects of type String. The String class is located in the java.lang package, which is automatically imported by all programs by default.
Methods to create String objects:
String s1 = "java";
String s2 = new String();
String s3 = new String("Java");Immutability of Java strings
String objects cannot be modified after they are created , is immutable. The so-called modification actually creates a new object, pointing to a different memory space.
If you need a string that can be changed, you can use StringBuffer or StringBuilder.
Each time a new string is generated, a new object is generated. Even if the contents of the two strings are the same, it will be "false" when compared using "==". If you only need to compare whether the contents are the same, you should Use the "equals()" method.
The constant pool in Java is used to save data in compiled class files that have been determined during compilation.
package cn.test;public class Demo12 { public static void main(String[] args) {
String s1 = "java";//先检查字符串常量池中是否有"java"字符串，如果有则直接指向，如果没有就在字符串常量池中添加"java"字符串并指向它，所以这种方式创建字符串时最多创建一个对象，或者不创建对象
String s2 = "java";//s2直接指向字符串常量池中的"java"
String s3 = new String("java");//在堆内存申请一块内存存储字符串"java",s3指向其内存块对象，同时检查字符串常量池中是否有"java"字符串，如果没有就添加字符串"java"到常量池中，所以new String()有可能创建两个对象
String s4 = new String("java");
System.out.println(s1 == s2);
System.out.println(s1 == s3);
System.out.println(s3 == s4);
s1 = "欢迎来到" + s1;
System.out.println(s1);
System.out.println(s3.equals(s4));
}
}Execution result:
true
false
false
Welcome to java
true
- ##String class Commonly used methods
##Example 1:
String fileName = "HelloWorld.java";
String email = "xiaoli@163.com";
// 判断.java文件名是否正确：合法的文件名应该以.java结尾
int index = fileName.lastIndexOf('.');
String prefix = fileName.substring(index+1);
if ( index > 0 && prefix.equals("java")) {
System.out.println("Java文件名正确");
} else {
System.out.println("Java文件名无效");
}
// 判断邮箱格式是否正确：合法的邮箱名中至少要包含"@", 并且"@"是在"."之前
int index2 = email.indexOf('@');
int index3 = email.indexOf('.');
if (index2 != -1 && index3 > index2) {
System.out.println("邮箱格式正确");
} else {
System.out.println("邮箱格式无效");
}
String str = "boo:and:foo";
String[] arr = str.split(":");
for (int i = 0; i < arr.length; i++)
{
System.out.print(arr[i]);
}String str = "boo:and:foo";
String[] arr = str.split(":");
for
(int i = 0; i < arr.length; i++)
{
System.out.print(arr[i]);
}Execution results:
The Java file name is correct
The email format is correct booandfoo
Example 2:
String str = "abcd阿";
byte[] b = str.getBytes();
for (int j = 0; j < b.length; j++) {
System.out.print("[" + b[j] + "]");
}Running result:
[97][98][99][100][-80][- 94]
Note: 1 byte is equal to 8 bits. In gbk encoding, 1 Chinese character storage requires 2 bytes, and 1 English character storage requires 1 byte.
Example 3 :
String s = "aljlkdsflkjsadjfklhasdkjlflkajdflwoiudsafhaasdasd";
// 出现次数int num = 0;
// 循环遍历每个字符，判断是否是字符 a ，如果是，累加次数for ( int i = 0;i < s.length(); i++ )
{ // 获取每个字符，判断是否是字符a
if ( s.charAt(i) == 'a' ) { // 累加统计次数
num++;
}
}
System.out.println("字符a出现的次数：" + num);Running result:
Number of occurrences of character a: 8
- StringBuilder class
- The String class is mutable, and many temporary
will be generated when strings are frequently manipulated. This problem can be avoided by using StringBuilder or StringBuffer. They are basically similar, the difference is that StringBuffer is thread safe, so the performance is slightly higher. Therefore, in general, to create a string object with variable content, the StringBuilder class is preferred. StringBuilder str2 = new StringBuilder("java");
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