C++ random float

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

Take a look at Boost.Random. You could do something like this:

float gen_random_float(float min, float max)
{
    boost::mt19937 rng;
    boost::uniform_real<float> u(min, max);
    boost::variate_generator<boost::mt19937&, boost::uniform_real<float> > gen(rng, u);
    return gen();
}

Play around, you might do better passing the same mt19937 object around instead of constructing a new one every time, but hopefully you get the idea.

Answer 2

This will generate a number from 0.0 to 1.0, inclusive.

float r = (float)rand()/(float)RAND_MAX;

Note that the rand() function will often not be sufficient if you need truly random numbers.

Answer 3

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 4

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 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

I wasn't satisfied by any of the answers so far so I wrote a new random float function. It makes bitwise assumptions about the float data type. It still needs a rand() function with at least 15 random bits.

//Returns a random number in the range [0.0f, 1.0f).  Every
//bit of the mantissa is randomized.
float rnd(void){
  //Generate a random number in the range [0.5f, 1.0f).
  unsigned int ret = 0x3F000000 | (0x7FFFFF & ((rand() << 8) ^ rand()));
  unsigned short coinFlips;

  //If the coin is tails, return the number, otherwise
  //divide the random number by two by decrementing the
  //exponent and keep going. The exponent starts at 63.
  //Each loop represents 15 random bits, a.k.a. 'coin flips'.
  #define RND_INNER_LOOP() \
    if( coinFlips & 1 ) break; \
    coinFlips >>= 1; \
    ret -= 0x800000
  for(;;){
    coinFlips = rand();
    RND_INNER_LOOP(); RND_INNER_LOOP(); RND_INNER_LOOP();
    //At this point, the exponent is 60, 45, 30, 15, or 0.
    //If the exponent is 0, then the number equals 0.0f.
    if( ! (ret & 0x3F800000) ) return 0.0f;
    RND_INNER_LOOP(); RND_INNER_LOOP(); RND_INNER_LOOP();
    RND_INNER_LOOP(); RND_INNER_LOOP(); RND_INNER_LOOP();
    RND_INNER_LOOP(); RND_INNER_LOOP(); RND_INNER_LOOP();
    RND_INNER_LOOP(); RND_INNER_LOOP(); RND_INNER_LOOP();
  }
  return *((float *)(&ret));
}

Answer 7

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;