ternary operator doubt

[Nik Arora]

Hi All,
For the below code the output is compilation fails. Can anybody explain me why?


public class ObjectTest {

public static void main(String[] args) {
byte b = 0;
(true) ? b = b + 1 : b = 100;

System.out.println("Output: " + b);
}
}
Source: TeamTesting
Thanks All

[fred rosenberger]

have you tried compiling it? What does the error message say? That error message usually give a pretty good idea as to why the compilation fails.

[Chandra Bhatt]

Case #1: The leftside of the assignment must be variable. This is the reason behind compiler error.

One more thing (not part of compiler error until Case #1 is satisfied)

byte b1 =10;
b1 =b1+1; //error, can't cast from int to byte

byte + int results int, so you need explicit casting.



Thanks,

[Sanjay Singh]

 Case #1: The leftside of the assignment must be variable. This is the reason behind compiler error.
 
One more thing (not part of compiler error until Case #1 is satisfied)
 
byte b1 =10;
b1 =b1+1; //error, can't cast from int to byte
 
byte + int results int, so you need explicit casting.   

Hi Chandra, your comment for Case#1 might not correct. e.g see this code, it compiles -
byte b = 0;
b = (true) ? b = (byte) (b + 1) : b;
here it seems to be that we must provide a variable which holds the returned value(obviously we have to do type cast if return type is different).

Case # 2: is correct, but again it creates the problem, if i write above code like this-

  byte b = 0; 
b = (true) ? b = (byte) (b + 1) : b = (byte)50; 
  

It gives compiler error. Don't know why?

[Chandra Bhatt]

Case #1: The leftside of the assignment must be variable. This is the reason behind compiler error.

[Sanjay]:

Hi Chandra, your comment for Case#1 might not correct. e.g see this code, it compiles -
byte b = 0;
b = (true) ? b = (byte) (b + 1) : b;

That was compiler generated error, saying the left side of the assignment must be variable that is what you did:

The form of ternary operator is:

boolean expression? if true : else part

BTW,
try this:

b = (true) ? b = (byte) (b + 1)   :(   b = (byte)50); //compiles fine

[ May 08, 2007: Message edited by: Chandra Bhatt ]

[swarna dasa]

Assignment operator has least precedence.

Try
b = (true) ? (b = (byte) (b + 1)) : (b = (byte)50);

[Nik Arora]

Assignment operator has least precedence.

Try
b = (true) ? (b = (byte) (b + 1)) : (b = (byte)50);

Hi Swarna,
I have two doubts look at the below statements

byte b=0;
b = (true) ? b= (byte) (b + 1) : b= 50; //This code does not compile;

b=(true) ? b=(byte) (b+1) : (b=50); // This code compiles

My first doubt is when parantheses is given the code compiles why it does not compile without parantheses. Whats the difference in providing parantheses and not providing in the above statements?.


My second doubt look at the below statement

b=(true)

Is it a assignment statement?. If yes then b is byte how can we assign boolean to it.

[Nik Arora]

Assignment operator has least precedence.

Try
b = (true) ? (b = (byte) (b + 1)) : (b = (byte)50);

Hi Swarna,
I have two doubts look at the below statements

byte b=0;
b = (true) ? b= (byte) (b + 1) : b= 50; //This code does not compile;

b=(true) ? b=(byte) (b+1) : (b=50); // This code compiles

My first doubt is when parantheses is given the code compiles why it does not compile without parantheses. Whats the difference in providing parantheses and not providing in the above statements?.


My second doubt look at the below statement

b=(true)

Is it a assignment statement?. If yes then b is byte how can we assign boolean to it.

[swarna dasa]

Look at this code as well.

               boolean b1=false,b2=false,b3=false;
 
if(b1 && b2=false || b3) {
System.out.println("test");
}

Does this fail?

Anyway sticking to your question.
Question 1:
b=(true) ? b=(byte) (b+1) : (b=50);

You have a ternary operator when you say
b=(true) ? b=(byte) (b+1) : b=50;

It is seen by the compiler as follows:

b=((true) ? b=(byte) (b+1) : b)=50;
since it is a ternary operator
after ? the expression till : is evaluated as one expression and = is left out.

This expression works fine
b = true ? b = 50:b; //Though b=50 is not in paranthesis as ? till : is evaluated

Question 2:
b=(true)
No. Reaon you have ?, compiler knows that it is a ternary operator hence evaluates the condition ? and :.

[Meena R. Krishnan]

Nik wrote:
byte b=0;
b = (true) ? b= (byte) (b + 1) : b= 50; //This code does not compile;

b=(true) ? b=(byte) (b+1) : (b=50); // This code compiles

My first doubt is when parantheses is given the code compiles why it does not compile without parantheses. Whats the difference in providing parantheses and not providing in the above statements?.

In the case where there is no parenthesis, the expression is seen like this

b =((true)? b=(byte)(b+1):b) =50;

and therefore the error.

nik wrote:
My second doubt look at the below statement

b=(true)

Is it a assignment statement?. If yes then b is byte how can we assign boolean to it.

You are not assigning 'true' to b instead the result of the expresion.

Ternary operator:
variable = (condition) ? expression1 : expression2

Depending on whether the condition evaluates to a true or false, assign the value of expression1 or expression2 to the variable.

In your example the condition evaluates to True and therefore the first expression's result gets assigned to b.
[ May 08, 2007: Message edited by: M Krishnan ]

[debasmita pattnayak]

hi,
here is my doubt regarding the following code posted by swarna

code:
--------------------------------------------------------------------------------

boolean b1=false,b2=false,b3=false; if(b1 && b2=false || b3) {System.out.println("test");}

--------------------------------------------------------------------------------

The code doesnt compile.
Can anyone explain me why?

[Meena R. Krishnan]

boolean b1=false,b2=false,b3=false; if(b1 && b2=false || b3) { System.out.println("test"); }

It is because the compiler sees the above expression as (see the brackets enclosing (b1 && b2) )

 boolean b1=false,b2=false,b3=false; if([B]([/B]b1 && b2[B])[/B]=false || b3) { System.out.println("test"); }

and therefore considers '=false' as out of place.

Instead the following might work enclose ( b2=false) in brackets)

 boolean b1=false,b2=false,b3=false; if(b1 && [B]([/B]b2=false[B])[/B] || b3) { System.out.println("test"); }

[ May 08, 2007: Message edited by: M Krishnan ]

[Nik Arora]

Thanks Krishnan and Swarna

[Chandra Bhatt]

Originally posted by debasmita pattnayak:
hi,
here is my doubt regarding the following code posted by swarna

code:
--------------------------------------------------------------------------------

boolean b1=false,b2=false,b3=false; if(b1 && b2=false || b3) {System.out.println("test");}

--------------------------------------------------------------------------------

The code doesnt compile.
Can anyone explain me why?

Again, the issue of = operator, least priority operator;
&& operator finds both side of the operand boolean and does its job

b1 && b2 = false
then
false = false // error

So make it like this:
if(b1 && (b2=false) || b3) { System.out.println("test");

In this case expression inside () operator is evaluated first and then
everything goes fine.


Thanks,

[debasmita pattnayak]

hi everyone,
thanks for such beautiful explainations.
my doubt has been clarified.

[swarna dasa]

Posted by Krishnan

It is because the compiler sees the above expression as (see the brackets enclosing (b1 && b2) ) and therefore considers '=false' as out of place.

I think (b1 && b2) is wrong.

Posted by Chandra

&& operator finds both side of the operand boolean and does its job

b1 && b2 = false
then
false = false // error

Having && on the left side of the = operator is the problem.

I dont think we can have any operator to the left hand side of assignment operator.

e.g.

int i=0,j=0,k=0,l=0;
i=i+j=k+j;