Operators

[Swati Thorve]

Can anybody tell me the order of evaluation for string s

public class Question49 {
public static void main(String[] args) {
boolean b = false;
String s =
(b=!b)?(b=!b)?"Hello":"hello": ( b=!b)?"world":"World";
System.out.println(s);
}
}


The options are
A.Prints: Hello
B. Prints: hello
C. Prints: Helloworld
D. Prints: helloWorld
E. Compilation error.

Correct answer is B

Plzz explain this.

[ January 05, 2006: Message edited by: Swati Thorve ]

[ January 05, 2006: Message edited by: Swati Thorve ]
[ January 05, 2006: Message edited by: Swati Thorve ]

[Mehul Sanghvi]

String s =
(b=!b)?(b=!b)?"Hello":"hello": ( b=!b)?"world":"World";

The first (b=!b) is evaluated to true and hence the control comes to second (b=!b) which will now be evaluated to false and hence the value of "hello" will be assigned to String s.

A possible catch here is = vs. ==
The expn (b=!b) is actually an assignment operation rather then a conditional statement (which is generally expected in ternary operators).

Regards.

[Thomas De Vos]

Let's play compiler here ...


1. I'm adding brackets

    
String s = (b=!b) ? (b=!b) ? "Hello" : "hello" : (b=!b) ? "world" : "World";

2. I'm adding more brackets

String s = (b=!b) ? ( (b=!b) ? "Hello" : "hello")  :  ( (b=!b) ? "world" : "World");

3. I'm assigning !b to b

String s = true ? ( (b=!b) ? "Hello" : "hello" ) :  ( (b=!b) ? "world" : "World" );

4. I'm evaluating the first condition ...

String s = true ? ( (b=!b) ? "Hello" : "hello" ) :  ( (b=!b) ? "world" : "World" );
 
          |----------------------------|

5. I'm assigning !b to b again

String s = true ? ( false ? "Hello" : "hello" ) 

6. Got my result

s => "hello"

[Swati Thorve]

Thanks a lot Mehul

[Swati Thorve]

Thanks a lot Thomas,For such a wonderful explanation.Thank u very much