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This article brings you an introduction to the addition and comparison of String strings (detailed). It has certain reference value. Friends in need can refer to it. I hope it will be helpful to you. help.
String String addition and comparison
public static void main(String[] args) {
String a = "helloword";
final String b = "hello";
String d = "hello";
String c = b + "word";
String e = d + "word";
String f ="hello"+"word";
System.out.println(a == c);
System.out.println(a == e);
System.out.println(a == f);
}
输出： true false true
The first thing to make clear is the reference data type = = What is compared is the address value, equal is not rewritten, what is compared is the address value, after rewriting, what is compared is the content. String is rewritten, StringBuffer is not rewritten
Secondly:
a= =c is true because b is a final type. When the final variable is a basic data type or String type, if its exact value can be known during compilation, the compiler will use it as a compile-time constant. That is to say, where the final variable is used, it is equivalent to a direct access to the constant and does not need to be determined at runtime. Therefore, in the above piece of code, since the variable b is modified by final, it will be treated as a compiler constant, so where b is used, the variable b will be directly replaced with its value. So c = b "word" is equal to c = "hello" "word" is the same as ---> When final, the content is compared directly instead of the address.
a ==e is false because e = d "word" actually creates a StringBuffer object, and then uses the StringBuffer object to execute the append method to create the string object "ab", and then converts it to String . But this converted String object, that is, the s3 above, is placed in the heap. And s4 is a string constant, placed in the constant pool. So what is returned is false. ----->The address values are different
a ==f is true because: When the constants are added, they are actually added directly to "helloword" during compilation. This is the JVM's Optimized, so when running, the bytecodes of a and f are the same. Because there is a "helloword" in the constant pool, the two references point to a string "helloword", so the return result is also true. ----->Point to the same address
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