Question

Given: int num1, num2, newNum; double x, y; Which of the following assignments are valid? If an assignment is not valid, state the reason. When not given, assume that each variable is declared. a. num1 = 35; b. newNum = num1 – num2; c. num1 = 5; num2 = 2 + num1; num1 = num2 / 3; d. num1 * num2 = newNum; e. x = 12 * num1 - 15.3; f. num1 * 2 = newNum + num2; g. x / y = x * y; h. num2 = num1 % 2.0; i. newNum = static_cast (x) % 5; j. x = x + y - 5; k. newNum = num1 + static_cast (4.6 / 2);

Short answer

Assignments a, b, c, e, i, j, and k are valid.

Step-by-step solution

Analyze Assignment a

The statement `num1 = 35;` is valid. `num1` is an integer, and the value `35` is also an integer, so the assignment is appropriate.

Analyze Assignment b

The statement `newNum = num1 - num2;` is valid. Both `num1` and `num2` are integers, and `newNum` is also an integer, which is suitable for storing the result of the subtraction of two integers.

Analyze Assignment c

The assignment `num1 = 5; num2 = 2 + num1; num1 = num2 / 3;` is valid. Each sub-assignment correctly uses integers, and integer division is used.

Analyze Assignment d

The statement `num1 * num2 = newNum;` is invalid. The left side of the assignment is an expression, not a variable, and expressions cannot be used as l-values in assignments.

Analyze Assignment e

The statement `x = 12 * num1 - 15.3;` is valid. The expression evaluates to a double because `15.3` is a double, which can be stored in `x`.

Analyze Assignment f

The statement `num1 * 2 = newNum + num2;` is invalid. Similar to assignment d, the left side is an expression and cannot be used as an l-value.

Analyze Assignment g

The statement `x / y = x * y;` is invalid. The left side is an expression, and expressions, which are not variables, cannot be on the left side of the assignment.

Analyze Assignment h

The statement `num2 = num1 % 2.0;` is invalid. The modulus operator `%` cannot be used with double values, such as `2.0`. Both operands must be integers.

Analyze Assignment i

The statement `newNum = static_cast (x) % 5;` is valid. `x` is cast to an integer, which makes using the `%` operator with `5` (an integer) correct.

Analyze Assignment j

The statement `x = x + y - 5;` is valid. All terms are expressions of type double, and they can be assigned to `x`.

Analyze Assignment k

The statement `newNum = num1 + static_cast (4.6 / 2);` is valid. The result of `4.6 / 2` is cast to an integer before being added to `num1`, which is also an integer.

Key concepts

Integer and Double Data Types

In C++, variables can have different data types, which dictate not only what values they can hold but also how operations on them are handled and stored. Two common types are `int` (integer) and `double` (double-precision floating point).

• **Integers (`int`)**: Whole numbers without any fractional part. They are efficient in storing numeric values for counts, indexes, and discrete metric values.
• **Doubles (`double`)**: Numbers that include a fractional part. They are appropriate for representing continuous measurements, such as heights, weights, or calculations involving decimals.

It is crucial to understand these distinctions because mixing types can lead to implicit type conversions, which may not always be desired and can sometimes cause precision loss.

Assignment Validity

In C++, the validity of an assignment depends largely on the type matching and the structure of the assignment statement. Each assignment needs to be evaluated for compatibility between the variable on the left-hand side (LHS) and the value or expression on the right-hand side (RHS).

• A simple assignment like `num1 = 35;` is valid when the data type of the RHS is appropriate for the LHS variable.
• An expression assigned to a variable, such as `newNum = num1 - num2;`, must result in a value that matches the variable’s type on the LHS.
• Expressions themselves, like `num1 * num2`, cannot be on the LHS of an assignment; only actual variables can be assigned values.

Careful checking of assignments ensures that errors due to incompatibilities and misuse of expressions as assignment targets are avoided.

Modulus Operator Usage

The modulus operator `%` is used in C++ to find the remainder of the division of two integers. It is important to remember that both operands must be integers for the operation to be valid.
For example, `num2 = num1 % 2;` correctly uses the modulus operator because both `num1` and `2` are integers.
However, attempting to use a float or double, like `num1 % 2.0`, results in a syntax error since the modulus operator is not defined for floating point types. When performing operations that might involve fractional components, the `%` operator should be limited to integer types only.

static_cast in C++

Type casting in C++ lets you explicitly convert a variable from one data type to another. `static_cast` is a type-safe option that prevents some type-related errors by ensuring that the operation is legal at compile time.
For example, `newNum = static_cast(x) % 5;` first converts `x` from a `double` to an `int` before using the modulus operator. This conversion is necessary because the modulus operation requires both operands to be integers.
`static_cast` is often preferred over other casting methods like C-style casts because it provides a clear, precise way of handling type conversions with built-in checks.