到目前为止，我对任何答案都不满意，所以我写了一个新的随机浮点函数。它对浮点数据类型进行了按位假设。它仍然需要一个rand()函数，至少有15个随机位。

//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));
}

rand()可用于在c++中生成伪随机数。结合RAND_MAX和一点数学运算，您可以在任意选择的间隔内生成随机数。这对于学习目的和玩具程序来说是足够的。如果需要真正具有正态分布的随机数，则需要使用更高级的方法。

这将生成一个从0.0到1.0的数字。

float r = static_cast <float> (rand()) / static_cast <float> (RAND_MAX);

这将生成一个从0.0到任意浮点数X的数字:

float r2 = static_cast <float> (rand()) / (static_cast <float> (RAND_MAX/X));

这将生成一个从任意LO到任意HI的数字:

float r3 = LO + static_cast <float> (rand()) /( static_cast <float> (RAND_MAX/(HI-LO)));

注意，如果需要真正的随机数，rand()函数通常是不够的。

在调用rand()之前，必须首先通过调用srand()来“播种”随机数生成器。这应该在程序运行期间执行一次，而不是在每次调用rand()时执行一次。通常是这样做的:

srand (static_cast <unsigned> (time(0)));

为了调用rand或srand，你必须#include <cstdlib>。

为了调用time，你必须#include <ctime>。

调用带有两个浮点值的代码，代码可以在任何范围内工作。

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

如果您知道您的浮点数格式是IEEE 754(几乎所有现代cpu，包括Intel和ARM)，那么您可以使用逐位方法从一个随机整数构建一个随机浮点数。只有当你无法访问c++ 11的random或Boost时，才应该考虑这样做。随机的，两者都更好。

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;
}

这将比使用除法得到更好的分布。

对于c++，它可以在dist变量指定的范围内生成实浮点数

#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);
}

#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));
...

我不能给出两个答案，所以这里是第二个答案。Log2随机数，很大程度上偏向于0.0f但它实际上是一个随机浮动1.0f到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));
...