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Suppose a, b, and sum are int variables and \(\mathrm{c}\) is a double variable. What value is assigned to each variable after each statement executes? Suppose \(\mathbf{a}=3\) $$\begin{array}{l} b=5, \text { and } c=14.1 & \text { ___ }& \text { ___ }& \text { ___ }& \text { ___ }\\\ \text { sum } =a+b+c & \text { ___ }& \text { ___ }& \text { ___ }& \text { ___ }\\\ c /=a & \text { ___ }& \text { ___ }& \text { ___ }& \text { ___ }\\\ a *=2 * b+c ;& \text { ___ }& \text { ___ }& \text { ___ }& \text { ___ } \\\\\end{array}$$

Short Answer

Expert verified
Final values: a = 14, b = 5, c = 4.7, sum = 22.1.

Step by step solution

01

Initial Values

Starting with the initial values, we have- \(a = 3\)- \(b = 5\)- \(c = 14.1\).
02

Calculate Sum

The sum is calculated by adding \(a\), \(b\), and \(c\):\[\text{sum} = a + b + c = 3 + 5 + 14.1 = 22.1\]After this step, \( \text{sum} = 22.1 \).
03

Division Operation

The variable \(c\) is divided by \(a\):\[c = \frac{c}{a} = \frac{14.1}{3} = 4.7\]Now, \(c = 4.7\).
04

Update a with Multiplication and Addition

We update \(a\) using the expression \(a = 2 * b + c\):\[a = 2 * 5 + 4.7 = 10 + 4.7 = 14.7\]\(a\) gets the value \(14.7\). Remembering that \(a\) was an integer and therefore loses precision, the variable will actually be stored as \(a = 14\).
05

Final Values

After all operations:- \(a = 14\)- \(b = 5\)- \(c = 4.7\)- \(\text{sum} = 22.1\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Data Types
In C++ programming, data types are essential because they essentially define what kind of data a variable can hold. Each data type consumes a different amount of memory and provides different capabilities:
  • int: This is a basic data type used to store integers, like -3, 0, 123. It typically occupies 4 bytes of memory.
  • double: This data type is used for storing floating-point numbers, for example, 3.14159 or -0.001. It occupies more memory, usually 8 bytes, but it allows for decimals and fractions.
Data types help the compiler to understand how to process the data the program operates on. They also define the limits of your variables, ensuring you don't store a floating-point number in an int variable. Understanding data types leads to more efficient and error-free code.
Variables
Variables act as containers for storing data values in programming. They have a specific name and data type connected to them, and their values can change while code executes:
  • In the exercise, variables like a, b, c, and sum are initialized to hold different values.
  • Each of these variables gets assigned a type: int for whole numbers and double for numbers with decimals.
When you create a variable, you decide the name and the type. Using descriptive names makes code easier to read and maintain, while choosing the right data type ensures you can store the correct amount of information.
Arithmetic Operations
Arithmetic operations are the basic operations you can perform on numbers in C++, separated into addition, subtraction, multiplication, and division:
  • Addition (+): Combines numbers with each other, like \( a + b + c \).
  • Subtraction (-): Calculates the difference between numbers.
  • Multiplication (*): Multiplies numbers, important in expressions like \( 2 * b + c \).
  • Division (/): Divides one number by another; remember, when dividing integers, the result is an integer too.
Understanding the basic arithmetic operations unlocks the ability to solve mathematical expressions in your code. Always remember that in C++, integer arithmetic differs from floating point. Integer operations round down the result, like in dividing 7 by 2, which equals 3.
Type Casting
Type casting is the conversion of a variable from one data type to another. It's essential for ensuring data types match in operations:
  • Explicit type casting is when we manually convert, using syntax like (double)myInt.
  • Implicit type casting occurs automatically, such as when a double is assigned to an int variable - causing loss of precision.
In the original exercise, the variable a was supposed to be a float value due to its calculation, but was stored as an integer. This implicit type casting loses detail. Understanding type casting helps prevent unintentional data losses and makes sure you precisely control how data changes types.

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Most popular questions from this chapter

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). $$\begin{array}{l} a=(b++)+3 i & \text {___} & \text {___} & \text {___}\\\ c=2+a+(++b) i & \text {___} & \text {___} & \text {___}\\\ b=2 *(++c)-(a++) i & \text {___} & \text {___} & \text {___}\end{array}$$

Give meaningful identifiers for the following variables. a. A variable to store the first name of a student. b. A variable to store the discounted price of an item. c. A variable to store the number of juice bottles. d. A variable to store the number of miles traveled. e. A variable to store the highest test score.

What action must be taken before a variable can be used in a program?

include #include using namespace std; const int PRIME_NUM = 11; int ma… # What is printed by the following program? Suppose the input is: Miller 34 340 #include #include using namespace std; const int PRIME_NUM = 11; int main() { const int SECRET = 17; string name; int id; int num; int mysteryNum; cout << "Enter last name: "; cin >> name; cout << endl; cout << "Enter a two digit number: "; cin >> num; cout << endl; id = 100 * num + SECRET; cout << "Enter a positive integer less than 1000: "; cin >> num; cout << endl; mysteryNum = num * PRIME_NUM - 3 * SECRET; cout << "Name: " << name << endl; cout << "Id: " << id << endl; cout << "Mystery number: " << mysteryNum << endl; return 0; }

Which of the following are valid C++ assignment statements? Assume that i, x, and percent are double variables. a. i = i + 5; b. x + 2 = x; c. x = 2.5 *x; d. percent = 10%;

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