# Chapter 2 – Looping and Counting

Solutions to exercises in Chapter 2 of Accelerated C++, “Looping and counting.”

## Exercise 2-1

Change the framing program so that it writes its greeting with no separation from the frame.

### Solution

The simplest way to accomplish this is to change the value of the constant `pad` to `0`.

An alternative is to remove the `pad` constant from the program.

The output of each program is as follows:

## Exercise 2-2

Change the framing program so that it uses a different amount of space to separate the sides from the greeting than it uses to separate the top and bottom borders from the greeting.

### Solution

In the program below the `pad` constant has been removed, and in its place are now two constants, `vertical_pad` and `horizontal_pad`. The `vertical_pad` constant is used to calculate the number of rows to write, and the `horizontal_pad` constant is used to calculate the width of each line written.

The output of the program is as follows:

## Exercise 2-3

Rewrite the framing program to ask the user to supply the amount of spacing to leave between the frame and the greeting.

### Solution

This solution is similar to the first solution provided for exercise 2-1, except that it
asks the user to input the amount of padding to display inside the frame. Note that the `pad` variable is no
longer constant since it must be determined when the program runs.

The output of the program is as follows:

## Exercise 2-4

The framing program writes the mostly blank lines that separate the borders from the greeting one character at a time. Change the program so that it writes all the spaces needed in a single output expression.

### Solution

This solution creates a `string` constant named `padding` that consists of the appropriate number of spaces to output between the border character and the greeting. It also adds a `filler` constant that consists of the number of spaces between the borders on lines above and below the greeting.

In the output loop, after determining that spaces should be output, the program tests to determine if the current row is a padding row or the greeting row, and outputs the appropriate `string` constant.

The output of the program is as follows:

## Exercise 2-5

Write a set of “*” characters so that they form a square, a rectangle, and a triangle.

### Solution

This solution outputs each of the three figures to be `height` rows. Changing this constant will change the size of each shape. Each shape is built until `height` rows have been printed.

The output of the program is as follows:

## Exercise 2-6

What does the following code do?

### Solution

This code counts from 1 to 10, displaying each number on a new line as it counts.

## Exercise 2-7

Write a program to count down from `10` to `-5`.

### Solution

The loop variable starts at 11, which is outside the beginning of the range. The program decrements the loop variable until it is equal to the end of the range.

The output of the program is as follows:

## Exercise 2-8

Write a program to generate the product of the numbers in the range `[1,10)`.

### Solution

The half-open range `[1,10)` includes the numbers 1 through 9. The program loops until all the numbers in the range have been multiplied. The variable `product` contains the product of the range when the loop exits.

The output of the program is as follows:

## Exercise 2-9

Write a program that asks the user to enter two numbers and tells the user which number is larger than the other.

### Solution

The program asks the user for two numbers, then tests if the first number is less than the second number. If the first number is the larger number, the program reports that fact; otherwise, it reports that the second number is the larger number.

This program does not handle the situation where the two numbers are equal.

The output of the program is as follows:

## Exercise 2-10

Explain each of the uses of `std::` in the following program:

### Solution

The first use of `std::` is in the `using`-declaration. The name `cout` is declared to mean `std::cout`. The second time `std::` appears is to make use of the `std::cout` stream to output a blank line. Because the `using`-declaration does not apply at this line, the `std::` namespace must be used explicitly. The last use of `std::` is to output a new line with the `std::endl` manipulator.

## 9 thoughts on “Chapter 2 – Looping and Counting”

1. Snoopz says:

Exercise 2-1:
I for one understood the exercise as I was asked only to not separate the statement from the frame. Not to remove the blanks altogether. Because if you removed the blanks there would be no reason for { // are we on the border?
if (r == 0 || r == rows – 1 ||
c == 0 || c == cols – 1)
cout << "*";
else
cout < the original being
–> const string::size_type cols = greeting.size() +pad*2 + 2;
and
—>if ( r == pad + 1 && c == 1 ) –> starting code was –> if ( r == pad + 1 && c == pad + 1 )
We check if we have written the first ‘*’ and if we are right in the middle ready to write the greeting..

This was the way I understood the exercise. It might not be correct but then there’s a redundant amount of code if we aren’t writing any empty lines.

2. Snoopz says:

Somehow I cannot get the full code to show :(

1. I’ll edit your comment a little later on when I get time. I’m awfully swamped at the moment. :(

3. Syl says:

I used a different method for the triangle of exercise 2-5, I used(for) instead of (while). I don’t know if you would consider this less or more complicated, however, I wanted to share with anyone would like to see.

Note: I personnaly like to ask the user to input the size even though it wasn’t asked in the exercise, if any was wondering why I added that

//ask how big the triagle shall be

cout << "how high do you want the triangle to be" <> rows;
cout << endl;

//Determine number of colums

const int cols = rows * 2 – 1;

// Start drawing the tringle
//invariant : we have r rows so far

for (int r = 0; r != rows ; ++r){

//invariant: we have c columns so far
for (int c = 0; c != cols; ++c){
if (r == rows -1) // This if draws the base
cout << "*";
else
if (c == (cols/2 – r) || c == (cols/2 + r) ) //could also be c == rows -1 +r and c == rows – 1 + r
cout << "*";
else
cout << " ";
}

cout << endl;

}

1. Syl says:

hmm, its seems to no properly show the code I wrote.

first part should be

//ask how big the triagle shall be

cout << "how high do you want the triangle to be" <> rows;
cout << endl;

//Determine number of colums

const int cols = rows * 2 – 1;

1. Syl says:

it did it again, I give up, sry for the spam

4. Taylan says: