Question

Given that x = 2, y = 1, z = 0, what will the following cout statement display?

cout << "answer = " << (x || !y && z) << endl;
a. answer = 0
b. answer = 1
c. answer = 2
d. None of these

Answer 1

The cout statement will display "answer = 1" because in the expression (x || !y && z), x being non-zero causes the entire expression to evaluate to true (1) when using the logical OR operator.

Step-by-step explanation:

The cout statement in question involves using logical operators (||, !, and &&) on integer values. In C++, the logical OR operator (||) returns true if either operand is non-zero, the logical NOT operator (!) inverts the boolean value of its operand (true becomes false and vice versa), and the logical AND operator (&&) returns true only if both operands are true (non-zero in this context).

Let's evaluate the expression (x || !y && z) step by step:

1. x || anything will always be true (which is represented as 1 in C++) because x is 2, which evaluates to true.
2. The logical NOT operator (!) applied to y (which is 1) will give false (0) because !1 is 0.
3. The result of !y ANDed with z doesn't matter because x || has already guaranteed the result to be true.
4. Therefore, the entire expression (x || !y && z) evaluates to true (1).

So the cout statement displays: answer = 1.