C++11 随机数
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C++11 random numbers
提问by Baruch
I need to generate random numbers, but from as wide a range as possible (64 bit at least). I don't care if the distribution is perfect, so std::rand()would work, but it only returns an int. I understand that c++11 has some random number generating capability that can give any size number, but is very complex to use. Can someone post a simple example of how to use it as simply as possible to get the described functionality (64 bit or more random numbers) in as simple a way as possible (like std::rand())?
我需要生成随机数，但范围尽可能大（至少 64 位）。我不在乎分布是否完美，所以std::rand()会起作用，但它只返回一个int. 我知道 c++11 有一些随机数生成能力，可以给出任何大小的数字，但使用起来非常复杂。有人可以发布一个简单的示例，说明如何尽可能简单地使用它来以尽可能简单的方式（例如std::rand()）获得所描述的功能（64 位或更多随机数）？
回答by jogojapan
This is how to use the C++11 random number generation for this purpose (adjusted from http://en.cppreference.com/w/cpp/numeric/random/uniform_int_distribution):
这是为此目的使用 C++11 随机数生成的方法（从http://en.cppreference.com/w/cpp/numeric/random/uniform_int_distribution调整）：
#include <random>
#include <iostream>
int main()
{
/* Initialise. Do this once (not for every
random number). */
std::random_device rd;
std::mt19937_64 gen(rd());
/* This is where you define the number generator for unsigned long long: */
std::uniform_int_distribution<unsigned long long> dis;
/* A few random numbers: */
for (int n=0; n<10; ++n)
std::cout << dis(gen) << ' ';
std::cout << std::endl;
return 0;
}
Instead of unsigned long long, you could use std::uintmax_tfrom cstdintto get the largest possible integer range (without using an actual big-integer library).
代替unsigned long long，您可以使用std::uintmax_tfromcstdint来获得最大可能的整数范围（不使用实际的大整数库）。
回答by kennytm
We could easily wrap a random number generator engine into srand/rand-like methods like this:
我们可以很容易地将随机数生成器引擎包装成类似 srand/rand 的方法，如下所示：
#include <random>
#include <iostream>
struct MT19937 {
private:
static std::mt19937_64 rng;
public:
// This is equivalent to srand().
static void seed(uint64_t new_seed = std::mt19937_64::default_seed) {
rng.seed(new_seed);
}
// This is equivalent to rand().
static uint64_t get() {
return rng();
}
};
std::mt19937_64 MT19937::rng;
int main() {
MT19937::seed(/*put your seed here*/);
for (int i = 0; i < 10; ++ i)
std::cout << MT19937::get() << std::endl;
}
(Like srandand rand, this implementation does not care about thread-safety.)
（就像srandand rand，这个实现不关心线程安全。）
Well the wrapper functions are so trivial that you could just use the engine directly.
好吧，包装函数非常简单，您可以直接使用引擎。
#include <random>
#include <iostream>
static std::mt19937_64 rng;
int main() {
rng.seed(/*put your seed here*/);
for (int i = 0; i < 10; ++ i)
std::cout << rng() << std::endl;
}
回答by RiaD
Not C++11, but easy enough
不是 C++11，但足够简单
((unsigned long long)rand() << 32) + rand()Here we generate two parts of int64 as int32's
((unsigned long long)rand() << 32) + rand()这里我们将int64的两部分生成为int32的
As JasonDpointed out, it assumes that rand()generate 32bit integer. It' spossible to xor
rand() << x, rand() << (2*x), rand() << (3*x), etc, where x<= bit's in generate by rand()number`. It should be OK too.
正如所JasonD指出的，它假设rand()生成 32 位整数。它可以 xor
rand() << x、rand() << (2*x)、rand() << (3*x)等，其中x<= 位在按rand()数字生成。也应该没问题。
