Question

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;
    }

Answer 1

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.