String classString

String is a special wrapper class data.

public final class String
    implements java.io.Serializable.Comparable<String>, CharSequence
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String = new String(” ABC “); String STR =” ABC “; Is created in the form of

Common methods in strings

Why is String set to immutable

Mainly from the following three aspects:

1. Ensure the security of String objects. Assuming strings are mutable, strings can be maliciously modified.

2. Ensure that the hash attribute value will not change frequently to ensure uniqueness, so that the corresponding key-value caching function can be realized by containers like HashMap.

String constant pool can be implemented

String storage in the JVM

Code sample

   String str1 = "123";
   String str2 = "123";
   String str3 = "123";
   String str4 = new String("123");
   String str5 = new String("123");
   String str6 = new String("123"); Str1 = = str2:trueStr2 = = str3:trueStr3 = = str4:falseStr4 = = str5:falseStr5 = = str6:false
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JVM storage examples

Process for creating a String object

For the underlying JVM, String STR = new String(“123”) what is the process for creating an object?

1. Check the constant pool for the presence of the string “123”; If yes, the corresponding reference instance is returned. If no, create the corresponding instance object.
2. New a String object of type “123” in the heap;
3. Copy the object address to STR, and then create an application.

Note: If there is no string “123” in the constant pool, two objects are created; If it exists, an object and its reference are created.

Creating a String object

Question:

1.String str =”ab” + “cd”; Number of objects?

Analysis: if the string constant pool the string object

1. String constant pool :(1 object) “abcd”;
2. Heap: no
3. Stack: (1 reference) STR total: 1 object +1 reference

2.String str = new String(“abc”); Number of objects?

Analysis: if the string constant pool the string object

1. String constant pool :(1 object) “ABC “;

2. Heap :(1 object) new String(” ABC “)

3. Stack :(1 reference) STR

   Total: 2 objects +1 reference

3.String str = new String(“a” + “b”); Number of objects?

Analysis: if the string constant pool the string object

1. String constant pool :(3 objects) “a”, “b”, “ab”;
2. Heap :(1 object) new String(“ab”)
3. Stack: (1 reference) STR total: 4 objects +1 reference

4.String str = new String(“ab”) + “ab”; Number of objects?

Analysis: if the string constant pool the string object

1. String constant pool :(1 object) “ab”;
2. Heap :(1 object) new String(“ab”)
3. Stack: (1 reference) STR total: 2 objects +1 reference

5.String str = new String(“ab”) + new String(“ab”); Number of objects?

Analysis: if the string constant pool the string object

1. String constant pool :(1 object) “ab”;
2. Heap (2 objects) New String(“ab”), new String(“ab”)
3. Stack: (1 reference) STR total: 3 objects +1 reference

6.String str = new String(“ab”) + new String(“cd”); Number of objects?

Analysis: if the string constant pool the string object

1. String constant pool :(2 objects) “ab”, “CD “;
2. Heap :(2 objects) new String(“ab”), new String(” CD “)
3. Stack: (1 reference) STR total: 4 objects +1 reference

7.String str3 = str1 + str2; Number of objects?

String str1 = "ab";
String str2 = "cd";
String str3 = str1 + str2;
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Analysis: if the string constant pool the string object

1. String constant pool :(2 objects) “ab”, “CD”, “abcd”;
2. Heap: no
3. Stack: (3 references) STR1, STR2, STR3 Total: 2 objects +3 references

String objects are specified using the java.lang.string.intern () method

code

String str1 = "123";
String str2 = new String("123");
String str3 = str2;
System.out.println(Str1 == str2: + (str1 == str2));
System.out.println("str1 == str3：" + (str1 == str3));

// Specify a String object with the java.lang.string.intern () method
String str4 = str2.intern();
System.out.println("str1 == str4：" + (str1 == str4));
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Results:

Str1 = = str2:falseStr1 = = str3:falseStr1 = = str4:true
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The length of the String

The maximum length of a Java String is divided into two phases, the compile phase and the runtime phase

Compilation stage:

When we define strings directly using String literals, we store a copy of the String in the constant pool. Each constant in the constant pool is a table with its own type. For the index, u2 is defined, which is 2 bytes unsigned. A CONSTANT_String_info table of fixed length is used to store literal String values. Note that this table stores only literal String values, not symbolic references.

The JVM’s constant pool can hold up to 65535 entries. The 0th term doesn’t have to. The last item can only be 65534(subscript value) at most. The maximum length of a utF-8 constant string in each entry is 65535 bytes (not characters).

Runtime phase:

A String is stored internally as a char array whose length is int. The maximum length allowed for a String is integer. MAX_VALUE, which is 2147483647. And since characters in Java are stored in 16 bits, it takes about 4GB of memory to store the maximum length of a string.

String Often meet test questions

1. How to compare two strings? Equals or equals

In simple terms, “==” checks whether two references refer to the same object in memory, and equals() checks whether the values of two objects are equal. Whenever you want to check whether two strings are equal, you have to use equals().

2. Why is char[] better for security-sensitive String information than String?

Strings are immutable, which means they won’t change until the garbage collector comes along to clean them up. With arrays, you can explicitly modify character elements anywhere in it. That way, security-sensitive information such as passwords cannot appear anywhere on the system.

3. Can string objects be used in switch expressions?

As of JDK7, we can use strings in switch conditional expressions, which is not possible prior to version 7.

// java 7 only!
switch (str.toLowerCase()) {
    case "a":
        value = 1;
        break;
    case "b":
        value = 2;
        break;
}
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4. How to convert a string to an integer value?

int n = Integer.parseInt(“10”);

It is so simple that it is often used and occasionally forgotten.

5. How do I separate strings with Spaces?

We can easily use regular expressions to separate them. “\s” means space, as well as “, “\t “,” \r “, “\n”.

String[] strArray = aString.split(“\s+”);

6. What exactly does the subString() method do?

In JDK6, this method only gives a window to represent the resulting string on the character array that identifies the existing string, but does not create a new string object. If you need to create a new string object, you can add an empty string to the result like this:

str.subString(m, n) + “”

This creates a new character array to represent the resulting string. Also, there is a chance that your code will run faster because the garbage collector will recycle large strings that are not in use and leave behind substrings.

The subString() method in Oracle JDK7 creates a new character array instead of the existing one. Take a look at this figure to see the difference between the subString() method in JDK6 and JDK7.

7.String&StringBuilder&StringBuffer

String vs StringBuilder: StringBuilder is variable, and this means that you can also go to modify it after creating objects of value. StringBuilder vs StringBuffer: StringBuffer is synchronous, means it is thread-safe, but some will be slower than the StringBuilder.

8. How to quickly repeat a string?

In Python, you can just multiply a string by a number. In Java, you can use the repeat() method from the StringUtils class in the Apache Commons Lang package.

String str = “abcd”; String repeated = StringUtils.repeat(str,3); //abcdabcdabcd

9. How to convert a time-formatted string into a date object?

String str = “Sep 17, 2013”; Date date = new SimpleDateFormat(“MMMM d, yy”, Locale.ENGLISH).parse(str); System.out.println(date); //Tue Sep 17 00:00:00 EDT 2013

10. How do I count the number of occurrences of a character in a string?

You can do this using the StringUtils class in the Apache Commons Lang package.

int n = StringUtils.countMatches(“11112222”, “1”); System.out.println(n);