A profound understanding of String
1) String a string constant: the String object Once you create the object can not be changed. (Source below)
String str1 = "abc";
String str2 = "abc";
String str3 = new String("abc");
System.out.println(str1 == str2);
System.out.println(str1 == str3);
operation result
true
false
- == Comparison of basic data types when the comparison is a value, comparing the time reference type is the address value comparison. Here is a reference to the type of comparison.
- Creating "abc" first-time constant pool to see if there is not, then you create one, any direct use. Therefore, the same address value stored str1 and str2, point is the same.
- Because str3 out of new objects, out of the new stack, and the constant pool str1, the address values can not be equal, it is false.
2）How many Objects created with: String str=new String(“abc”)？
String str=new String("abc");
A: I create two objects in a heap in a constant pool
- The implementation of "abc" after a creation, new at the time to create a constant pool in the heap, and the constant pool "abc" a copy of the past. Then assigns a reference to the s1.
3) supplement Case
①
String s1 = "a" + "b" + "c";// 在编译时就变成 abc 常量池中创建abc
String s2 = "abc";
System.out.println(s1 == s2);// true java中有常量优化机制
System.out.println(s1.equals(s2));// true
At compile time becomes constant pool abc create abc, both in the constant pool
②
String s1 = "ab";
String s2 = "abc";
String s3 = s1 + "c";
System.out.println(s3 == s2);// false
System.out.println(s3.equals(s2));// true
Because this is equivalent to new s3 out, the corresponding address in the stack, s2 corresponding addresses in the constant pool
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2020.02.25
