CODEWITHSUNDEEP
javacompiler-errorsconditional-statements
Jimmy
Jimmy

Reputation: 2619

Compilation error in if-condition

At the very last condition, I was expecting to get the condition evaluated starting from the expression on the left of operator "&&" which becomes true and then the statement on the right . However, I am getting compilation error here saying "syntax error on token "=", != expected". Any explanation on this problem would be of great help.

boolean boolTrue=Boolean.TRUE;
    boolean assignBool=Boolean.FALSE;
    int ten=10;

    //eventually evaluates to true , runs okay
    if(assignBool=boolTrue){
        System.out.println("Executed");
    }
    //evaluates to true && true: runs correctly
    if( assignBool=boolTrue && ten==10 )
        System.out.println("Executed");

    //evaluates to true && true : runs correctly
    if( ten==10 && (assignBool=boolTrue)  )
        System.out.println("Executed");

    /*was supposed to evaluate to true && true : but gives compile error
    compiler expects me to use '!=' or '==' operator though the second statement ultimately evaluates to true 
    as in above case*/
    if( ten==10 && assignBool=boolTrue  )//Compile error
        System.out.println("Executed");

EDIT: Thanks for the answer. To check if it was operatar precedence issue, I ran same case in for loop and YES, that was it . 

    boolean boolTrue=Boolean.TRUE;
    boolean assignBool=Boolean.TRUE;
    int ten=10;

    singleloop:
    for(int i=0;((ten==10 && assignBool)==boolTrue);System.out.println("Executed")){
        i++;
        if(i>10)
            break singleloop;
    }

Upvotes: 3

Views: 635

Answers (5)

mpen
mpen

Reputation: 282805

it's being evaluated as if((ten==10 && assignBool)=boolTrue ) which means you are trying to assign to a constant.

Upvotes: 1

Ted Hopp
Ted Hopp

Reputation: 234795

The compiler is trying to evaluate the last if as if it were written

if( (ten==10 && assignBool)=boolTrue  )

which makes no sense because the left side is not an lvalue (to borrow some vocabulary from C). This is because && has higher precedence than = in Java.

Upvotes: 5

CrazyCasta
CrazyCasta

Reputation: 28302

I assume from your code that you are trying to play around and actually want to assign boolTrue to assignBool.

The problem is the order of operators. The && has higher precedence than = (source). It is creating a bool by anding with the ten==10 expression. The result is a non-lvalue (an expression that cannot be written to). Your expression is equivalent to the fully parenthesized:

((ten == 10) && assignBool) = boolTrue

As you've already found out, the following is the way to do it:

ten==10 && (assignBool=boolTrue)

Upvotes: 3

meriton
meriton

Reputation: 70564

As you can see from the nice table in the Java Tutorial, the assignment operator has the lowest precedence among all operators. That is, the expression

ten==10 && assignBool=boolTrue

is evaluated as

(ten==10 && assignBool) = boolTrue

but the left hand expression is not assignable.

If you want different precedence, you must use brackets, i.e.

ten==10 && (assignBool = boolTrue)

Upvotes: 3

chris
chris

Reputation: 412

I don't know java but it's probably binding && more strongly than your other operators.

Upvotes: 2

Related Questions