return (n > 2) ? n = 5 : n = 4; doesn't work?

Question

Why does this return an error

return (n > 2) ? n = 5 : n = 4;

but this does not

return (n > 2) ? n = 5 : n + 4;

Should it not be able to return n depending on either case?

Answer 1

The code you have can't be compiled because the ternary operator has a higher operator precedence than the assignment operator:

Operator Precedence

1. postfix ( expr++ expr-- )
2. unary ( ++expr --expr +expr -expr ~ ! )

...

13. ternary ( ? : )
14. assignment ( = += -= *= /= %= &= ^= |= <<= >>= >>>= )

When parsing the code

(n > 2) ? n = 5 : n = 4;

it will parse it like this:

(n > 2) ? n = 5 : n     // ternary operator
                    = 4 // assignment operator

This would result in pseudo code like this:

someResultValue 
                = value;

That will not work and you get compile errors like:

'Syntax error on token "=", <= expected'

or

'Type mismatch: cannot convert from Object & Comparable<?> & Serializable to int'.

You can use parentheses to let java see the third argument of the ternary operator as n = 4 . The code would look like this:

return (n > 2) ? n = 5 : (n = 4);

Answer 2

I think you may have any syntax error. I tried that in chrome JS console and worked fine!!

Update: Some people argued in comments that I didn't write function so I wrote in functional format and got the same result!

Answer 3

syntax: condition? expression1: expression2;

Try this

public class Solution {
    public static void main(String[] args) throws IOException {
       int a = 10;
       int b = (a>5)?a=5:0;
       System.out.println("a = "+ a);
       System.out.println("b = "+b);
       
    }
}

output: a = 5 b = 5