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);