Generate random numbers uniformly over an entire range

Why rand is a bad idea

Most of the answers you got here make use of the rand function and the modulus operator. That method may not generate numbers uniformly (it depends on the range and the value of RAND_MAX), and is therefore discouraged.

C++11 and generation over a range

With C++11 multiple other options have risen. One of which fits your requirements, for generating a random number in a range, pretty nicely: std::uniform_int_distribution. Here's an example:

const int range_from  = 0;
const int range_to    = 10;
std::random_device                  rand_dev;
std::mt19937                        generator(rand_dev());
std::uniform_int_distribution<int>  distr(range_from, range_to);

std::cout << distr(generator) << '\n';

And here's the running example.

Template function may help some:

template<typename T>
T random(T range_from, T range_to) {
    std::random_device                  rand_dev;
    std::mt19937                        generator(rand_dev());
    std::uniform_int_distribution<T>    distr(range_from, range_to);
    return distr(generator);
}

Other random generators

The <random> header offers innumerable other random number generators with different kind of distributions including Bernoulli, Poisson and normal.

How can I shuffle a container?

The standard provides std::shuffle, which can be used as follows:

std::vector<int> vec = {4, 8, 15, 16, 23, 42};

std::random_device random_dev;
std::mt19937       generator(random_dev());

std::shuffle(vec.begin(), vec.end(), generator);

The algorithm will reorder the elements randomly, with a linear complexity.

Boost.Random

Another alternative, in case you don't have access to a C++11+ compiler, is to use Boost.Random. Its interface is very similar to the C++11 one.

Warning: Do not use rand() for statistics, simulation, cryptography or anything serious.

It's good enough to make numbers look random for a typical human in a hurry, no more.

See Jefffrey's reply for better options, or this answer for crypto-secure random numbers.

Generally, the high bits show a better distribution than the low bits, so the recommended way to generate random numbers of a range for simple purposes is:

((double) rand() / (RAND_MAX+1)) * (max-min+1) + min

Note: make sure RAND_MAX+1 does not overflow (thanks Demi)!

The division generates a random number in the interval [0, 1); "stretch" this to the required range. Only when max-min+1 gets close to RAND_MAX you need a "BigRand()" function like posted by Mark Ransom.

This also avoids some slicing problems due to the modulo, which can worsen your numbers even more.

The built-in random number generator isn't guaranteed to have a the quality required for statistical simulations. It is OK for numbers to "look random" to a human, but for a serious application, you should take something better - or at least check its properties (uniform distribution is usually good, but values tend to correlate, and the sequence is deterministic). Knuth has an excellent (if hard-to-read) treatise on random number generators, and I recently found LFSR to be excellent and darn simple to implement, given its properties are OK for you.

If RAND_MAX is 32767, you can double the number of bits easily.

int BigRand()
{
    assert(INT_MAX/(RAND_MAX+1) > RAND_MAX);
    return rand() * (RAND_MAX+1) + rand();
}

If you are able to, use Boost. I have had good luck with their random library.

uniform_int should do what you want.

If you are concerned about randomness and not about speed, you should use a secure random number generation method. There are several ways to do this... The easiest one being to use OpenSSL's Random Number Generator.

You can also write your own using an encryption algorithm (like AES). By picking a seed and an IV and then continuously re-encrypting the output of the encryption function. Using OpenSSL is easier, but less manly.

You should look at RAND_MAX for your particular compiler/environment. I think you would see these results if rand() is producing a random 16-bit number. (you seem to be assuming it will be a 32-bit number).

I can't promise this is the answer, but please post your value of RAND_MAX, and a little more detail on your environment.

Check what RAND_MAX is on your system -- I'm guessing it is only 16 bits, and your range is too big for it.

Beyond that see this discussion on: Generating Random Integers within a Desired Range and the notes on using (or not) the C rand() function.

Using a Mersenne Twister engine (C++11):

#include <random>

// Returns a random integer within the range [min, max]
int generateRandomInt(const int min, const int max) {
  static bool is_seeded = false;
  static std::mt19937 generator;

  // Seed once
  if (!is_seeded) {
    std::random_device rd;
    generator.seed(rd());
    is_seeded = true;
  }

  // Use a Mersenne Twister engine to pick a random number
  // within the given range
  std::uniform_int_distribution<int> distribution(min, max);
  return distribution(generator);
}

If you want numbers to be uniformly distributed over the range, you should break your range up into a number of equal sections that represent the number of points you need. Then get a random number with a min/max for each section.

As another note, you should probably not use rand() as it's not very good at actually generating random numbers. I don't know what platform you're running on, but there is probably a better function you can call like random().

By their nature, a small sample of random numbers doesn't have to be uniformly distributed. They're random, after all. I agree that if a random number generator is generating numbers that consistently appear to be grouped, then there is probably something wrong with it.

But keep in mind that randomness isn't necessarily uniform.

The solution given by man 3 rand for a number between 1 and 10 inclusive is:

j = 1 + (int) (10.0 * (rand() / (RAND_MAX + 1.0)));

In your case, it would be:

j = min + (int) ((max-min+1) * (rand() / (RAND_MAX + 1.0)));

Of course, this is not perfect randomness or uniformity as some other messages are pointing out, but this is enough for most cases.

This is the solution I came up with:

#include "<stdlib.h>"

int32_t RandomRange(int32_t min, int32_t max) {
    return (rand() * (max - min + 1) / (RAND_MAX + 1)) + min;
}

This is a bucket solution, conceptually similar to the solutions that use rand() / RAND_MAX to get a floating point range between 0-1 and then round that into a bucket. However, it uses purely integer math, and takes advantage of integer division flooring to round down the value to the nearest bucket.

It makes a few assumptions. First, it assumes that RAND_MAX * (max - min + 1) will always fit within an int32_t. If RAND_MAX is 32767 and 32 bit int calculations are used, the the maximum range you can have is 32767. If your implementation has a much larger RAND_MAX, you can overcome this by using a larger integer (like int64_t) for the calculation. Secondly, if int64_t is used but RAND_MAX is still 32767, at ranges greater than RAND_MAX you will start to get "holes" in the possible output numbers. This is probably the biggest issue with any solution derived from scaling rand().

Testing over a huge number of iterations nevertheless shows this method to be very uniform for small ranges. However, it is possible (and likely) that mathematically this has some small bias and possibly develops issues when the range approaches RAND_MAX. Test it for yourself and decide if it meets your needs.

Of course, the following code won't give you random numbers, but pseudo random numbers.

Use the following code

#define QUICK_RAND(m,n) m + ( std::rand() % ( (n) - (m) + 1 ) )

For example:

int myRand = QUICK_RAND(10, 20);

You must call

srand(time(0));  // Initialize random number generator.

Otherwise the numbers won't be near random.

I just found this on the Internet. This should work:

DWORD random = ((min) + rand()/(RAND_MAX + 1.0) * ((max) - (min) + 1));