How to generate 64 bit random numbers?

问题

I'm implementing universal hashing and using the following universal hash function :

h(k)=((A*k)mod 2^64) rsh 64-r

where A is a random number between

2^61 and 2^62.

The rand() function in C++ has return type integer and it can't generate that big numbers. So how can i generate random numbers in this range? (numbers should be very random i.e. every number should have equal probability to be selected)

Note:

long long int random=rand();

doesn't work as the number returned by rand is int.

回答1:

In C++11 you can use the random header and std::uniform_int_distribution along with a 64-bit instance of std::mersenne_twister_engine this should do what you want (see it live):

#include <iostream>
#include <random>
#include <cmath>

int main()
{
    std::random_device rd;

    std::mt19937_64 e2(rd());

    std::uniform_int_distribution<long long int> dist(std::llround(std::pow(2,61)), std::llround(std::pow(2,62)));

    std::cout << std::llround(std::pow(2,61)) << std::endl; 
    std::cout << std::llround(std::pow(2,62)) << std::endl; 

    for (int n = 0; n < 10; ++n) {
            std::cout << dist(e2)<< ", " ;
    }
    std::cout << std::endl ;
}

If C++11 is not an option then it seems there is source code available for several 64-bit Mersenne Twister implementations.

回答2:

((long long)rand() << 32) | rand()

EDIT: that's assuming that rand() produces 32 random bits, which it might not.

来源：https://stackoverflow.com/questions/21096015/how-to-generate-64-bit-random-numbers