String concatenation and comparison gives unexpected result in println statement

问题

I couldn't figure out the following behaviour,

String str1= "abc";
String str2 = "abc";

System.out.println("str1==str2 "+ str1==str2);
System.out.println("str1==str2 " + (str1==str2))

Output for the above statement is as follows:

false

str1==str2 true

Why is this happening? Why the output is not like follows:

str1==str2 true

str1==str2 true

回答1:

+ has higher precedence than ==.
So your code :

System.out.println("str1==str2 " + str1 == str2);

will effectively be

System.out.println(("str1==str2 "+str1) == str2); 

so, you get false.

In case-2

System.out.println("str1==str2 " + (str1==str2));

you have used braces explicitly to compare str1 with str2 (which is true) and then append the value.

回答2:

The argument passed to println is evaluated left to right.

Therefore "str1==str2 "+ str1 concatenates two Strings, which are later compared to str2 and return a boolean.

回答3:

It's because of operator precedence.

In the first statement the + operator is executed before the == and "str1==str2 " is appended to str1, after which the result of the appending is compared with == to str2.

In the second statement the brackets () denote the atomic pieces that should be evaluated before the top-level operators (i.e. the +) take place. This is why first str1 is compared to str2 with ==, and then the result (true) is appended as a String to the "str1==str2 "

回答4:

System.out.println("str1==str2 "+ str1==str2);

in the above line, compiler checks if

"str1==str2"+str1

that is >>> "str1==str2 str1" is equal to str2 or not

That's why it prints it as false

回答5:

Comparing two String should always be done with the equals method. otherwise you compare if it is the same reference, not the same value!

来源：https://stackoverflow.com/questions/33914808/string-concatenation-and-comparison-gives-unexpected-result-in-println-statement