The code will not compile because in line 7, the line will work only if we use (x==y) 

in the line. The == operator compares values to produce a boolean, whereas the = 

operator assigns a value to variables.

Option A, B, and D are incorrect because the code does not get as far as compiling. 

If we corrected this code, the output would be false.

Option A is correct. The >>> operator moves all bits to the right, zero filling the left bits. 

The bit transformation looks like this:

Before: 1000 0000 0000 0000 0000 0000 0000 0000

After: 0000 0000 0000 0000 0000 0000 0000 0001

Option C is incorrect because the >>> operator zero fills the left bits, which in this case 

changes the sign of x, as shown.

Option B is incorrect because the output method print() always displays integers in base 10.

Option D is incorrect because this is the reverse order of the two output numbers.

None.

When the fix() method is first entered, start()'s s1 and fix()'s s1 reference variables both 

refer to the same String object (with a value of "slip"). Fix()'s s1 is reassigned to a new 

object that is created when the concatenation occurs (this second String object has a 

value of "slipstream"). When the program returns to start(), another String object is created, 

referred to by s2 and with a value of "stream".

None.

The left shift operator shifts all of the bite in a value to the left
specified number of times. For each

shift left, the high order bit is
shifted out and lost, zero is brought in from right. When a left shift
is

applied to an integer operand, bits are lost once they are shifted
past the bit position 31.

The controlling condition of ternary operator must evaluate to boolean.

Order of precedence is (highest to lowest) a -> b -> c -> d.

Unary not operator, ~, inverts all of the bits of its operand in binary representation.