Let x be an integer which can take a value of 0 or 1. The statement if (x == 0) x =1 ; else x =0 ; is equivalent to which one of the following ?

Let x be an integer which can take a value of 0 or 1. The statement <code>if (x == 0) x =1 ; else x =0 ;</code> is equivalent to which one of the following?

1. x = 1 + x;

2. x = x - 1;

3. x = 1 - x;

4. x = 1 % x;

The correct answer is: x = 1 - x;

To understand why the statement <code>x = 1 - x;</code> is equivalent to <code>if (x == 0) x = 1; else x = 0;</code>, consider the logical flow of the given if-else statement:

• When <code>x == 0</code>: The statement <code>x = 1;</code> is executed.

• When <code>x == 1</code>: The statement <code>x = 0;</code> is executed.

Now, let's analyze each choice to see which one replicates this behavior:

1. <code>x = 1 + x;</code> <br>This does the following:

   • If <code>x = 0</code>, then <code>x = 1 + 0 = 1</code>.

   • If <code>x = 1</code>, then <code>x = 1 + 1 = 2</code>.

   Therefore, it does not correctly replicate the behavior of the given if-else statement.

2. <code>x = x - 1;</code> <br>This does the following:

   • If <code>x = 0</code>, then <code>x = 0 - 1 = -1</code>.

   • If <code>x = 1</code>, then <code>x = 1 - 1 = 0</code>.

   Again, this does not correctly replicate the behavior because <code>x</code> can become -1, which is not a valid outcome based on the original problem constraints (i.e., <code>x</code> can only be 0 or 1).

3. <code>x = 1 - x;</code> <br>This does the following:

   • If <code>x = 0</code>, then <code>x = 1 - 0 = 1</code>.

   • If <code>x = 1</code>, then <code>x = 1 - 1 = 0</code>.

   Hence, this choice perfectly replicates the behavior of the given if-else statement.

4. <code>x = 1 % x;</code> <br>This choice is problematic because it attempts to divide by zero when <code>x</code> is 0, which will result in a runtime error. Therefore, it is not a valid transformation of the given problem.

Therefore, upon examination of each choice, we conclude that the correct answer is: x = 1 - x;