Java : space makes a difference in compilation?

Question

I was making a program (A Piglatin sort of...), in which I unintentionally missed a variable in the statement:

String a = "R"++'a';

It should actually have been String a = "R"+text+'a';. The compiler produced an error. But, when I made it:

String a = "R"+ +'a';

The program compiled.

I am wondering why putting a space made the difference even though Java does not care whether you put a space or not in certain statements, like : String a="ABCD"; is the same as String a = "ABCD";

Can someone please explain this behavior?

Answer 1

++ is an operator in its own right (pre or post increment).

Putting it between a string and a char literal is not syntactically valid.

But with "R"+ +'a', the second + will bind to the char literal a and will act as the unary plus operator (this operator has a very high precedence). This is not a no-op: in Java it has the effect of promoting the type of a to an int. This type promotion means that the output will be R97 rather than Ra (97 is the ASCII number for a). The remaining + acts as the string concatenator.

Answer 2

Because ++ is a unary operator, while + + is interpret as following:

1. "R"+ represents a string concatenation and then the second argument (for the concatenation is evaluated)
2. +'a' represents an explicit setting of sign (+) to the (numeric) char literal 'a'. Because you've explicitly set the sign of the value, it is treated as a numeric one and hence the result of +'a' is the numeric representation of the character 'a' and that's why the result is R97.

The same way if you did String a = "R"+ -'a'; then the numeric value of 'a' (which is 97) would be negated and the result would be R-97.

If you, however, had just ignored the + sign before 'a', then 'a' would have been treated as character, not as numeric, and the result would be Ra.