#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));
...

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()返回一个介于0和RAND_MAX之间的int值。要获得0.0到1.0之间的随机数，首先将rand()返回的int转换为浮点数，然后除以RAND_MAX。

在一些系统上(目前想到的是带有VC的Windows)， RAND_MAX小得可笑，也就是只有15位。当除以RAND_MAX时，你只生成了一个15位的尾数，而不是23位。这对您来说可能是问题，也可能不是问题，但在这种情况下，您会遗漏一些值。

哦，刚才注意到已经有关于这个问题的注释了。不管怎样，这里有一些代码可以帮你解决这个问题:

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

未经测试，但可能工作:-)

#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));
...

在现代c++中，你可以使用c++11附带的<random>头文件。 要获得随机浮点数，可以使用std::uniform_real_distribution<>。

你可以使用一个函数来生成数字，如果你不希望数字总是相同的，那就将引擎和分布设置为静态。 例子:

float get_random()
{
    static std::default_random_engine e;
    static std::uniform_real_distribution<> dis(0, 1); // rage 0 - 1
    return dis(e);
}

理想的做法是将浮动对象放置在std::vector:这样的容器中:

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 << " ";
}

示例输出:

0.0518757 0.969106 0.0985112 0.0895674 0.895542