Question

Let’s see the following simplest code snippet in Java.

final public class Parsing
{
    public static void main(String[] args)
    {
        int p=10;
        int q=10;

        System.out.println(p==q);
    }
}

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The above code in Java is fine and displays true as both p and q of the same type (int) contain the same value (10). Now, I want to concatenate the argument of println() so that it contains a new line escape sequence \n and the result is displayed in a new line as below.

System.out.println("\n"+p==q);

Which is not allowed at all because the expression p==q evaluates a boolean value and a boolean type in Java (not Boolean, a wrapper type) can never be converted to any other types available in Java. Therefore, the above mentioned statement is invalid and it issues a compile-time error. What is the way to get around such situations in Java?

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and surprisingly, the following statements are perfectly valid in Java.

System.out.println("\n"+true);
System.out.println("\n"+false);

and displays true and false respectively. How? Why is the same thing in the statement System.out.println("\n"+p==q); not allowed?

Answer 1

You have to add parentheses () around p==q (the way you write it, it will be interpreted as (“\n”+p) == q, and String cannot be compared to a boolean). This operator precedence is desired for expressions like

if(a+b == c+d)

etc. So,

System.out.println("\n"+(p==q));

Should work just fine.