Random float number generation - c++

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

rand() can be used to generate pseudo-random numbers in C++. In combination with RAND_MAX and a little math, you can generate random numbers in any arbitrary interval you choose. This is sufficient for learning purposes and toy programs. If you need truly random numbers with normal distribution, you'll need to employ a more advanced method.
This will generate a number from 0.0 to 1.0, inclusive.
float r = static_cast <float> (rand()) / static_cast <float> (RAND_MAX);
This will generate a number from 0.0 to some arbitrary float, X:
float r2 = static_cast <float> (rand()) / (static_cast <float> (RAND_MAX/X));
This will generate a number from some arbitrary LO to some arbitrary HI:
float r3 = LO + static_cast <float> (rand()) /( static_cast <float> (RAND_MAX/(HI-LO)));
Note that the rand() function will often not be sufficient if you need truly random numbers.
Before calling rand(), you must first "seed" the random number generator by calling srand(). This should be done once during your program's run -- not once every time you call rand(). This is often done like this:
srand (static_cast <unsigned> (time(0)));
In order to call rand or srand you must #include <cstdlib>.
In order to call time, you must #include <ctime>.

Answer 2

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) ) ) ;
}

Answer 3

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 4

In modern c++ you may use the <random> header that came with c++11.
To get random float's you can use std::uniform_real_distribution<>.
You can use a function to generate the numbers and if you don't want the numbers to be the same all the time, set the engine and distribution to be static.
Example:
float get_random()
{
static std::default_random_engine e;
static std::uniform_real_distribution<> dis(0, 1); // rage 0 - 1
return dis(e);
}
It's ideal to place the float's in a container such as std::vector:
int main()
{
std::vector<float> nums;
for (int i{}; i != 5; ++i) // Generate 5 random floats
nums.emplace_back(get_random());
for (const auto& i : nums) std::cout << i << " ";
}
Example output:
0.0518757 0.969106 0.0985112 0.0895674 0.895542

Answer 5

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 6

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 7

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 8

drand48(3) is the POSIX standard way. GLibC also provides a reentrant version, drand48_r(3).
The function was declared obsolete in SVID 3 but no adequate alternative was provided so IEEE Std 1003.1-2013 still includes it and has no notes that it's going anywhere anytime soon.
In Windows, the standard way is CryptGenRandom().

Answer 9

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 10

In my opinion the above answer do give some 'random' float, but none of them is truly a random float (i.e. they miss a part of the float representation). Before I will rush into my implementation lets first have a look at the ANSI/IEEE standard format for floats:
|sign (1-bit)| e (8-bits) | f (23-bit) |
the number represented by this word is
(-1 * sign) * 2^e * 1.f
note the the 'e' number is a biased (with a bias of 127) number thus ranging from -127 to 126. The most simple (and actually most random) function is to just write the data of a random int into a float, thus
int tmp = rand();
float f = (float)*((float*)&tmp);
note that if you do float f = (float)rand(); it will convert the integer into a float (thus 10 will become 10.0).
So now if you want to limit the maximum value you can do something like (not sure if this works)
int tmp = rand();
float f = *((float*)&tmp);
tmp = (unsigned int)f // note float to int conversion!
tmp %= max_number;
f -= tmp;
but if you look at the structure of the float you can see that the maximum value of a float is (approx) 2^127 which is way larger as the maximum value of an int (2^32) thus ruling out a significant part of the numbers that can be represented by a float.
This is my final implementation:
/**
* Function generates a random float using the upper_bound float to determine
* the upper bound for the exponent and for the fractional part.
* #param min_exp sets the minimum number (closest to 0) to 1 * e^min_exp (min -127)
* #param max_exp sets the maximum number to 2 * e^max_exp (max 126)
* #param sign_flag if sign_flag = 0 the random number is always positive, if
* sign_flag = 1 then the sign bit is random as well
* #return a random float
*/
float randf(int min_exp, int max_exp, char sign_flag) {
assert(min_exp <= max_exp);
int min_exp_mod = min_exp + 126;
int sign_mod = sign_flag + 1;
int frac_mod = (1 << 23);
int s = rand() % sign_mod; // note x % 1 = 0
int e = (rand() % max_exp) + min_exp_mod;
int f = rand() % frac_mod;
int tmp = (s << 31) | (e << 23) | f;
float r = (float)*((float*)(&tmp));
/** uncomment if you want to see the structure of the float. */
// printf("%x, %x, %x, %x, %f\n", (s << 31), (e << 23), f, tmp, r);
return r;
}
using this function randf(0, 8, 0) will return a random number between 0.0 and 255.0

Answer 11

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 12

For C++, it can generate real float numbers within the range specified by dist variable
#include <random> //If it doesnt work then use #include <tr1/random>
#include <iostream>
using namespace std;
typedef std::tr1::ranlux64_base_01 Myeng;
typedef std::tr1::normal_distribution<double> Mydist;
int main() {
Myeng eng;
eng.seed((unsigned int) time(NULL)); //initializing generator to January 1, 1970);
Mydist dist(1,10);
dist.reset(); // discard any cached values
for (int i = 0; i < 10; i++)
{
std::cout << "a random value == " << (int)dist(eng) << std::endl;
}
return (0);
}

Answer 13

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 14

#include <cstdint>
#include <cstdlib>
#include <ctime>
using namespace std;
/* single precision float offers 24bit worth of linear distance from 1.0f to 0.0f */
float getval() {
/* rand() has min 16bit, but we need a 24bit random number. */
uint_least32_t r = (rand() & 0xffff) + ((rand() & 0x00ff) << 16);
/* 5.9604645E-8 is (1f - 0.99999994f), 0.99999994f is the first value less than 1f. */
return (double)r * 5.9604645E-8;
}
int main()
{
srand(time(NULL));
...
I couldn't post two answers, so here is the second solution. log2 random numbers, massive bias towards 0.0f but it's truly a random float 1.0f to 0.0f.
#include <cstdint>
#include <cstdlib>
#include <ctime>
using namespace std;
float getval () {
union UNION {
uint32_t i;
float f;
} r;
/* 3 because it's 0011, the first bit is the float's sign.
* Clearing the second bit eliminates values > 1.0f.
*/
r.i = (rand () & 0xffff) + ((rand () & 0x3fff) << 16);
return r.f;
}
int main ()
{
srand (time (NULL));
...