The code involves a series of bitwise and logical operations. Let's break it down step by step: 1) `int a = 2;` initializes a with 2. 2) `int b = ~a;` calculates the bitwise NOT of 2. Since 2 in binary is `0000 0010`, its bitwise NOT is `1111 1101`, which is -3 in decimal. Hence, `b = -3`. 3) `int c = a ^ b;` performs a bitwise XOR operation between 2 and -3. The result is -1, so `c = -1`. 4) `boolean d = a < b & a > c++;` uses bitwise AND. The expression `a < b` evaluates to `2 < -3`, which is false. The expression `a > c++` first checks `2 > -1`, which is true. Since bitwise AND with a false value results in false, `d` is false. After the evaluation, `c` is incremented to 0. Thus, the first print statement outputs `false 0`. 5) `boolean e = a > b && a > c++;` uses logical AND. The expression `a > b` evaluates to `2 > -3`, which is true. The expression `a > c++` checks `2 > 0`, which is true. Hence, `e` is true. After this evaluation, `c` is incremented to 1. Thus, the second print statement outputs `true 1`. Therefore, the result is `false 0` and `true 1`.