Week 1

Escape sequences

Exercise 1.1

Write a program that produces the output

"Hello"
'world'

Solution

#include <iostream>

using namespace std;

int main() {
    cout << "\"Hello\"\n\'world\'";
    return 0;
}

Alternative solution

#include <iostream>

using namespace std;

int main() {
    cout << "\"Hello\"" << endl << "\'world\'";
    return 0;
}

Variables

Exercise 1.2

Given the following program, fill in the blank with code that swaps the values of x and y (multiple statements may be required). Your code must work for all possible values of x and y.

#include <iostream>

using namespace std;

int main() {
    int x = 10;
    int y = 20;



    return 0;
}

Solution

We can introduce an intermediate variable z to store the old value of x as follows.

    int z = x;
    x = y;
    y = z;

Types

Numerical errors

Integer overflow may lead to unexpected results. For example, on many computers,

int i = 2147483647;
cout << i + 1;

produces the output -2147483648 since the integers ‘wrap around’ from the largest integer to the smallest integer.

Floating-point numbers may also overflow. For example,

double d = 1e308;
cout << 2 * d;

produces the output inf (representing $\infty$) since the largest double is approximately $1.8 \times 10^{308}$. Underflow is also possible with floating-point numbers:

double d = 1e-323;
cout << d / 4;

produces the output 0 since the smallest positive double is approximately $4.9 \times 10^{-324}$. In addition, floating-point numbers are susceptible to rounding errors. The statements

double d = 0.1 + 0.1 + 0.1 - 0.3;
cout << d;

give 5.55112e-17 since $0.1$ and $0.3$ cannot be exactly represented by a double. The size of the relative error in a floating-point arithmetic operation could be up to $\sim 10^{-16}$.

Operators

Exercise 1.3

What is the output of the following statements?

int x_int = 2;
int y_int = 3;
double x_double = 2.0;
double y_double = 3.0;

cout << x_int / y_int << '\n'; 
cout << x_double / y_int << '\n'; 
cout << x_int / y_double << '\n'; 
cout << x_double / y_double << '\n'; 

Solution

0
0.666667
0.666667
0.666667