Random float number generation

Question

How do I generate random floats in C++?

I thought I could take the integer rand and divide it by something, would that be adequate enough?

Answer 1

Call the code with two float values, the code works in any range.

float rand_FloatRange(float a, float b)
{
    return ((b - a) * ((float)rand() / RAND_MAX)) + a;
}

Answer 2

rand() return a int between 0 and RAND_MAX. To get a random number between 0.0 and 1.0, first cast the int return by rand() to a float, then divide by RAND_MAX.

Answer 3

If you know that your floating point format is IEEE 754 (almost all modern CPUs including Intel and ARM) then you can build a random floating point number from a random integer using bit-wise methods. This should only be considered if you do not have access to C++11's random or Boost.Random which are both much better.

float rand_float()
{
    // returns a random value in the range [0.0-1.0)

    // start with a bit pattern equating to 1.0
    uint32_t pattern = 0x3f800000;

    // get 23 bits of random integer
    uint32_t random23 = 0x7fffff & (rand() << 8 ^ rand());

    // replace the mantissa, resulting in a number [1.0-2.0)
    pattern |= random23;

    // convert from int to float without undefined behavior
    assert(sizeof(float) == sizeof(uint32_t));
    char buffer[sizeof(float)];
    memcpy(buffer, &pattern, sizeof(float));
    float f;
    memcpy(&f, buffer, sizeof(float));

    return f - 1.0;
}

This will give a better distribution than one using division.

Answer 4

If you are using C++ and not C, then remember that in technical report 1 (TR1) and in the C++0x draft they have added facilities for a random number generator in the header file, I believe it is identical to the Boost.Random library and definitely more flexible and "modern" than the C library function, rand.

This syntax offers the ability to choose a generator (like the mersenne twister mt19937) and then choose a distribution (normal, bernoulli, binomial etc.).

Syntax is as follows (shameless borrowed from this site):

  #include <iostream>
  #include <random>

  ...

  std::tr1::mt19937 eng;  // a core engine class 
  std::tr1::normal_distribution<float> dist;     

  for (int i = 0; i < 10; ++i)        
      std::cout << dist(eng) << std::endl;

Answer 5

On some systems (Windows with VC springs to mind, currently), RAND_MAX is ridiculously small, i. e. only 15 bit. When dividing by RAND_MAX you are only generating a mantissa of 15 bit instead of the 23 possible bits. This may or may not be a problem for you, but you're missing out some values in that case.

Oh, just noticed that there was already a comment for that problem. Anyway, here's some code that might solve this for you:

float r = (float)((rand() << 15 + rand()) & ((1 << 24) - 1)) / (1 << 24);

Untested, but might work :-)

Answer 6

C++11 gives you a lot of new options with random. The canonical paper on this topic would be N3551, Random Number Generation in C++11

To see why using rand() can be problematic see the rand() Considered Harmful presentation material by Stephan T. Lavavej given during the GoingNative 2013 event. The slides are in the comments but here is a direct link.

I also cover boost as well as using rand since legacy code may still require its support.

The example below is distilled from the cppreference site and uses the std::mersenne_twister_engine engine and the std::uniform_real_distribution which generates numbers in the [0,10) interval, with other engines and distributions commented out (see it live):

#include <iostream>
#include <iomanip>
#include <string>
#include <map>
#include <random>

int main()
{
    std::random_device rd;

    //
    // Engines 
    //
    std::mt19937 e2(rd());
    //std::knuth_b e2(rd());
    //std::default_random_engine e2(rd()) ;

    //
    // Distribtuions
    //
    std::uniform_real_distribution<> dist(0, 10);
    //std::normal_distribution<> dist(2, 2);
    //std::student_t_distribution<> dist(5);
    //std::poisson_distribution<> dist(2);
    //std::extreme_value_distribution<> dist(0,2);

    std::map<int, int> hist;
    for (int n = 0; n < 10000; ++n) {
        ++hist[std::floor(dist(e2))];
    }

    for (auto p : hist) {
        std::cout << std::fixed << std::setprecision(1) << std::setw(2)
                  << p.first << ' ' << std::string(p.second/200, '*') << '\n';
    }
}

output will be similar to the following:

0 ****
1 ****
2 ****
3 ****
4 *****
5 ****
6 *****
7 ****
8 *****
9 ****

The output will vary depending on which distribution you choose, so if we decided to go with std::normal_distribution with a value of 2 for both mean and stddev e.g. dist(2, 2) instead the output would be similar to this (see it live):

-6 
-5 
-4 
-3 
-2 **
-1 ****
 0 *******
 1 *********
 2 *********
 3 *******
 4 ****
 5 **
 6 
 7 
 8 
 9 

The following is a modified version of some of the code presented in N3551 (see it live) :

#include <algorithm>
#include <array>
#include <iostream>
#include <random>

std::default_random_engine & global_urng( )
{
    static std::default_random_engine u{};
    return u ;
}

void randomize( )
{
    static std::random_device rd{};
    global_urng().seed( rd() );
}

int main( )
{
  // Manufacture a deck of cards:
  using card = int;
  std::array<card,52> deck{};
  std::iota(deck.begin(), deck.end(), 0);

  randomize( ) ;  

  std::shuffle(deck.begin(), deck.end(), global_urng());
  // Display each card in the shuffled deck:
  auto suit = []( card c ) { return "SHDC"[c / 13]; };
  auto rank = []( card c ) { return "AKQJT98765432"[c % 13]; };

  for( card c : deck )
      std::cout << ' ' << rank(c) << suit(c);

   std::cout << std::endl;
}

Results will look similar to:

5H 5S AS 9S 4D 6H TH 6D KH 2S QS 9H 8H 3D KC TD 7H 2D KS 3C TC 7D 4C QH QC QD JD AH JC AC KD 9D 5C 2H 4H 9C 8C JH 5D 4S 7C AD 3S 8S TS 2C 8D 3H 6C JS 7S 6S

Boost

Of course Boost.Random is always an option as well, here I am using boost::random::uniform_real_distribution:

#include <iostream>
#include <iomanip>
#include <string>
#include <map>
#include <boost/random/mersenne_twister.hpp>
#include <boost/random/uniform_real_distribution.hpp>

int main()
{
    boost::random::mt19937 gen;
    boost::random::uniform_real_distribution<> dist(0, 10);

    std::map<int, int> hist;
    for (int n = 0; n < 10000; ++n) {
        ++hist[std::floor(dist(gen))];
    }

    for (auto p : hist) {
        std::cout << std::fixed << std::setprecision(1) << std::setw(2)
                  << p.first << ' ' << std::string(p.second/200, '*') << '\n';
    }
}

rand()

If you must use rand() then we can go to the C FAQ for a guides on How can I generate floating-point random numbers? , which basically gives an example similar to this for generating an on the interval [0,1):

#include <stdlib.h>

double randZeroToOne()
{
    return rand() / (RAND_MAX + 1.);
}

and to generate a random number in the range from [M,N):

double randMToN(double M, double N)
{
    return M + (rand() / ( RAND_MAX / (N-M) ) ) ;  
}