Can Someone please describe the process that happens in this code(line by line)?

Question

class Example{
public static void main(String args[]){
char a='a';
System.out.println(a=='\u0061');
System.out.println(\u0061=='\u0061');
System.out.println(\u0061==97);
\u0061='\u0041';
System.out.println('A'=='\u0041');
System.out.println(65=='\u0041');
System.out.println(65==a);
System.out.println('\u0041'==a);        
}
}

Output : true*7

I can't understand this code. Please help

If you replace every occurrence of `\u0061` in your code with `a`, and `\u0041` with `A`, it might make more sense to you. — khelwood, Feb 13 '19 at 10:36

khelwood · Answer 1 · 2019-02-13T11:00:37.757

According to the Java Language Specification:

A Unicode escape of the form \uxxxx, where xxxx is a hexadecimal value, represents the UTF-16 code unit whose encoding is xxxx.

The unicode value for the character 'a' is 97 (61 in hex), and for 'A' is 65 (41 in hex). So the character \u0061 in your source is read as a, and the character \u0041 is read as A.

Your code is read as:

class Example{
    public static void main(String args[]){
        char a='a';
        System.out.println(a=='a');
        System.out.println(a=='a');
        System.out.println(a==97);
        a='A';
        System.out.println('A'=='A');
        System.out.println(65=='A');
        System.out.println(65==a);
        System.out.println('A'==a);        
    }
}

Karol Dowbecki · Accepted Answer · 2019-02-13T10:59:24.277

char a='a'; - variable a stores a lowercase a character
System.out.println(a=='\u0061'); - in UTF lowercase a character is represented as \u0061
System.out.println(\u0061=='\u0061'); - \u0061 value is unquoted so it's decoded into lowercase a during compilation making it a=='\u0061', effectively the same as point 2.
System.out.println(\u0061==97); - same as above because 97dec = 61hex except here we are not using UTF notation to represent character, instead we use numerical value of char
\u0061='\u0041'; - a variable assigned value of \u0041 which is uppercase A
System.out.println('A'=='\u0041'); - same as point 2
System.out.println(65=='\u0041'); - 65dec = 41hex, same as point 4 but here we are comparing two constants, not the a variable.
System.out.println(65==a); - see above
System.out.println('\u0041'==a); - see above

"decoded into variable name during compilation" is misleading: it's decoded into the corresponding unicode symbol, `a`, which happens to be interpreted as a variable name in that specific context. — Andy Turner, Feb 13 '19 at 10:54

Can Someone please describe the process that happens in this code(line by line)?

2 Answers2