Question

Suppose \(a, b,\) and \(c\) are int variables and \(a=5\) and \(b=6 .\) What value is assigned to each variable after each statement executes? If a variable is undefined at a particular statement, report UND (undefined). a = (b++) + 3; __ __ __ c = 2 * a + (++b); __ __ __ b = 2 * (++c) - (a++);

Short answer

After executing all the statements, \(a = 10\), \(b = 45\), and \(c = 27\).

Step-by-step solution

Initialize Values

At the start, we have the following values: \(a = 5\) and \(b = 6\). The variable \(c\) is not defined initially and is therefore \(\text{UND}\).

Execute First Statement

The first statement is \(a = (b++) + 3\). Here, \(b++\) means the current value of \(b\) is used, and then \(b\) is incremented by 1 after the operation. So, initially \(b = 6\). Thus, the expression \((b++) + 3\) is \(6 + 3 = 9\). After this, \(b\) becomes 7. Now, \(a = 9\) and \(b = 7\). \(c\) remains \(\text{UND}\).

Execute Second Statement

The second statement is \(c = 2 * a + (++b)\). Here, \(++b\) means \(b\) is incremented first, making \(b = 8\), then the incremented value is used in the calculation. Now compute the expression: \(2 * a + (++b) = 2 * 9 + 8 = 18 + 8 = 26\). Therefore, \(c = 26\). Now, \(a = 9\), \(b = 8\), and \(c = 26\).

Execute Third Statement

The third statement is \(b = 2 * (++c) - (a++)\). First, increment \(c\) by 1, making \(c = 27\). Then calculate: \(2 * 27 = 54\). For \(a++\), use \(a = 9\) in the calculation first, making it 9, and increment \(a\) after the operation. So the expression is \(54 - 9 = 45\). Therefore, \(b = 45\). After this, \(a = 10\), \(b = 45\), and \(c = 27\).

Key concepts

Variable Initialization

In C++, initializing a variable means assigning it an initial value. Initially, when you declare a variable, such as `int a;`, it holds an unspecified value. To avoid unexpected behavior, especially when performing operations, it's crucial to initialize your variables properly.
For example, in the exercise, variables `a` and `b` are initialized with values 5 and 6, respectively, like so:

• `int a = 5;`
• `int b = 6;`

However, the variable `c` is not immediately initialized. In this step, it is considered undefined (UND). Before use in calculations, always ensure all variables have been initialized with an appropriate value.

Increment Operators

Increment operators are used to increase the value of a variable by one. They come in two main types: post-increment (`x++`) and pre-increment (`++x`).

• With post-increment (`x++`), the current value of the variable is used in any expression first, and then it's incremented.
• Conversely, with pre-increment (`++x`), the increment operation occurs first, and the updated value is returned and used in the expression.

In our example:

• In the expression `a = (b++) + 3`, `b++` provides the value 6, used immediately, and `b` becomes 7 afterward.
• In `c = 2 * a + (++b)`, `++b` increments `b` from 7 to 8, and then 8 is used in the expression.

Choosing the correct type of increment operator is critical to getting the desired results.

Arithmetic Operations

Arithmetic operations in C++ are fundamental for performing calculations. Common operations include addition, subtraction, multiplication, and division.

• Addition (`+`) is used to sum values, as seen in `a = (b++) + 3`.
• Multiplication (`*`) multiplies two values together, for instance, `2 * a` in the calculation of `c`.
• Complex expressions often combine these operations. In `c = 2 * a + (++b)`, multiplication and addition are used to compute the final result of 26.

Operator precedence plays a crucial role, where multiplication is performed before addition unless parentheses dictate another order. Understanding how these operations work allows for more efficient and error-free coding.

Variable Assignment

Variable assignment means storing a value (result of an expression) in a variable. The assignment operator (`=`) is used in C++, where what's on the right side of the `=` is computed and assigned to the variable on the left.
Let’s break down some assignments from the exercise:

• `a = 9;` - The result from `(b++) + 3` is stored in `a`.
• `c = 26;` - After calculating `2 * a + (++b)`, the result is assigned to `c`.
• `b = 45;` - The expression `2 * (++c) - (a++)` yields 45, which is then stored in `b`.

Variable assignment is a straightforward yet powerful concept, as it allows capturing and reusing results from expressions throughout your program.