11.3. Vector of random numbers

The first step is to generate a large number of random values and store them in a vector. By “large number,” of course, I mean 20. It’s always a good idea to start with a manageable number, to help with debugging, and then increase it later.

The following function takes a single argument, the size of the vector. It allocates a new vector of ints, and fills it with random values between 0 and upper_bound.

std::vector<int> make_vector (int size, int upper_bound) {
  std::vector<int> data (size);
  std::random_device r;
  std::default_random_engine eng(r());
  for (int& value: data) {
    value = std::uniform_int_distribution<int> {0, upper_bound} (eng);
  }
  return data;
}

The return type is std::vector<int>, which means that this function returns a vector of integers. To test this function, it is convenient to have a function that outputs the contents of a vector.

void print (const std::vector<int>& data) {
  for (const int& value: data) {
    cout << value << ' ';
  }
}

Notice that it is legal to pass vectors by reference. In fact it is quite common, since it makes it unnecessary to copy the vector. Since print does not modify the vector, we declare the parameter const. We use the same technique to get values out of the vector without copying every single one. Avoiding these kinds of copy operations makes code run faster. It is also less error prone.

The following code generates a vector and outputs it:

int num_values = 20;
int upper_bound = 9;
std::vector<int> numbers = make_vector (num_values, upper_bound);
print (numbers);

On my machine the output is

1 9 9 5 9 5 5 2 8 6 2 7 1 0 5 6 2 7 5 6

which is pretty random-looking. Your results might be different.

Try running the active code below!

#include <iostream> #include <random> #include <vector> // forward declarations std::vector<int> make_vector (int, int); void print (const std::vector<int>&); int main() { int num_values = 20; int upper_bound = 9; std::vector<int> numbers = make_vector (num_values, upper_bound); print (numbers); } std::vector<int> make_vector (int size, int upper_bound) { std::vector<int> data (size); std::random_device r; std::default_random_engine eng(r()); for (int& value: data) { value = std::uniform_int_distribution<int> {0, upper_bound} (eng); } return data; } void print (const std::vector<int>& data) { for (const int& value: data) { std::cout << value << ' '; } }

If these numbers are really random, we expect each digit to appear the same number of times—twice each. In fact, the number 5 appears five times, and the numbers 3 and 4 never appear at all.

Do these results mean the values are not really uniform? It’s hard to tell. With so few values, the chances are slim that we would get exactly what we expect. But as the number of values increases, the outcome should be more predictable.

To test this theory, we’ll write some programs that count the number of times each value appears, and then see what happens when we increase num_values.

Q-2: How should we declare the parameter, vector, if we don’t intend to make any changes to it?

Q-3: As we store more and more random numbers in a vector, we expect its contents to be __________.

• more uniform
• Correct!
• less uniform
• Incorrect! As we store more random numbers in a vector, we see that the frequencies of each number are approximately equal.
• more normal
• Incorrect! The distribution of random numbers is not related to the normal distribution.
• less normal
• Incorrect! The distribution of random numbers is not related to the normal distribution.

Q-4: Would compiling the following code lead to a compiler error?

void dostuff (const std::vector<int> & vec) {
   for (size_t i = 0; i < vec.size(); ++i) {
      vec[i] = i;
   }
}

• yes we would get a compile error
• Correct! we can't make changes to a vector we take in by constant reference
• no we would not because values are valid.
• Even if the values are legal and valid we are editing a constant which is not allowed.

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