Question

Suppose \(\mathbf{x}, \mathbf{y},\) and \(\mathbf{z}\) are int variables and \(\mathbf{w}\) and \(\mathbf{t}\) are double variables. What value is assigned to each of these variables after the last statement executes? x = 23; y = 35; x = x + y / 4 - 3; z = x % 3; w = 28 / 3 + 6.5 * 2; t = x / 4.0 + 15 % 4 - 3.5;

Short answer

x=28, z=1, w=22.3333, t=6.5.

Step-by-step solution

Calculate Expression for x

Initially, we have the integer variables set as:\( x = 23 \) and \( y = 35 \). The expression given for x is \( x = x + y / 4 - 3 \). Solve for \( y / 4 \), which results in integer division yielding \( 35 / 4 = 8 \) (since division of integers results in an integer, discarding the remainder). Thus, the expression for \( x \) becomes:\[ x = 23 + 8 - 3 = 28 \]. Therefore, after this step, \( x = 28 \).

Calculate Modulus for z

Next, we calculate the value of \( z \) using the expression: \( z = x \% 3 \).With the updated value of \( x = 28 \), we have:\[ z = 28 \% 3 = 28 - (3 \times 9) = 28 - 27 = 1 \].So, \( z = 1 \) after this step.

Evaluate Expression for w

The variable \( w \) is a double, so it uses a floating-point calculation: \( w = 28 / 3 + 6.5 * 2 \). First, calculate \( 28 / 3 \), which is approximately \( 9.3333 \), and \( 6.5 \times 2 = 13 \). Add these two results:\[ w = 9.3333 + 13 = 22.3333 \]. Therefore, \( w = 22.3333 \).

Evaluate Expression for t

Finally, calculate \( t \) with the expression: \( t = x / 4.0 + 15 \% 4 - 3.5 \). - First, \( x / 4.0 = 28 / 4.0 = 7.0 \), because dividing by a double results in a double.- Then, \( 15 \% 4 = 3 \), since the remainder of dividing 15 by 4 is 3.- Finally, combine these with \( - 3.5 \):\[ t = 7.0 + 3 - 3.5 = 6.5 \].Thus, \( t = 6.5 \).

Key concepts

Integer Division

When working with integer division in C++, you perform division where both the dividend and the divisor are integers. In such cases, C++ will discard any fractional part of the result and only return the whole number part.
For example, when you divide 35 by 4 using integer division, as seen in the problem, the result is 8 because 4 can fit into 35 eight complete times. The remainder, which is 3, is not included in the result here.

• Use integer division when you need no remainders.
• Keep in mind, it truncates towards zero, so be careful with this behavior in loops or index calculations.

Modulus Operation

The modulus operation, often represented by the symbol \%\, gives the remainder of the division of two numbers. It is primarily used with integers. This operation comes in handy when you need to determine if a number is divisible by another.
In the problem, the expression \( z = x \% 3 \) assigns the remainder of 28 divided by 3 to \( z \). Since 28 divided by 3 is 9 with a remainder of 1, \( z \) becomes 1.

• Perfect for checking even or odd numbers using \% 2.
• Use for circular data structures or repeating patterns where wrap-around is needed.

Floating Point Arithmetic

In floating point arithmetic, calculations can involve fractional numbers, and the precision is much higher compared to integers. The data type typically used in C++ is `double` or `float`.
For variables \( w \) and \( t \) in the given exercise, operations involve double calculations, causing the results to reflect properly with decimals. In \( w = 28 / 3 + 6.5 * 2 \), you see a result like 22.3333 showing full decimal representation.
Meanwhile, \( t = x / 4.0 + 15 \% 4 - 3.5 \) showcases precision using a floating point divisor to ensure decimals.

• Essential for scientific calculations where precision is critical.
• Always consider floating-point arithmetic anomalies, like precision errors when handling very large/small values.

Variable Assignment

Variable assignment in C++ allows you to store computed values in a variable for later use, improving code clarity and reusability. The assignment operator is the equal sign `=`.
During the exercise, various assignments are made, including \( x = 23 \), and later changing \( x \) to \( x + y / 4 - 3 \). This reflects how variable values can evolve through computation.
Similarly, operations with \( w \) and \( t \) showcase assignments of calculations involving both arithmetic and floating-point logic.

• Assign early for clarity and use comments to explain any complex logic embedded in calculations.
• Avoid assigning results of different types unless intentional, to prevent unexpected behavior.