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Standard rand() function gives numbers not big enough for me: I need unsigned long long ones. How do we get really big random numbers? I tried modifying a simple hash function but it's too big, takes too long to run and never produces numbers which are less than 1e5!!

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    How much "BIG" numbers you want? Commented Jan 23, 2015 at 17:50
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    How about using rand() to get sets of random bits, and populating each int-wide number of bits at a time into the unsigned long long using bit masks. Commented Jan 23, 2015 at 17:51
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    "never produces numbers which are less than 1e5" - how have you defined "never"? It would be expected that perfectly random numbers wouldn't occupy that range very often, given the full range you're asking for. Commented Jan 23, 2015 at 17:51
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    how about ditching rand() and using some of the better RNGs c++ provides en.cppreference.com/w/cpp/numeric/random Commented Jan 23, 2015 at 17:51
  • @ShauryaChats, VERY big, much more than 1e7 Commented Jan 23, 2015 at 17:54

7 Answers 7

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You can easily do this with std::uniform_int_distribution<unsigned long long>.

Simple example code (taken from here, modified to use unsigned long long):

#include <random> #include <iostream> int main() { std::random_device rd; std::mt19937 gen(rd()); std::uniform_int_distribution<unsigned long long> dis(lowerBorder, upperBorder); for (int n=0; n<10; ++n) std::cout << dis(gen) << ' '; std::cout << '\n'; } 

Note that the seeding of the mersenne twister as done here for demo purposes is not perfect, for example see here.

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3 Comments

That should be a correct answer. Also, it requires C++11 support.
I think you should use std::mt19937_64 for better randomness.
@Alexandru No, that will not improve randomness. You can try if it improves speed on your environment, that might potentially work.
11

Here's a portable C99 solution that returns a random 64-bit number:

unsigned long long llrand() { unsigned long long r = 0; for (int i = 0; i < 5; ++i) { r = (r << 15) | (rand() & 0x7FFF); } return r & 0xFFFFFFFFFFFFFFFFULL; } 

Explanation: rand() returns integers in the range 0 to RAND_MAX and RAND_MAX is only guaranteed to be at least 32,767 (15 random bits). long long is guaranteed to have 64 bits but may be larger.

13 Comments

Be really weary of this. It may seem reasonable but the numbers it produces may not (actually, probably will not) pass many statistical tests of randomness.
@AlterMann As I said, long long might be larger than 64 bits. If it is 64 bits, the compiler should make it a noop.
@LưuVĩnhPhúc MSVC (which has a reasonable market share) has #define RAND_MAX 0x7fff
@NikBougalis can you provide a citation or an example of a test that this would fail on (that rand() by itself would pass)
@NikBougalis Still not seeing issue; assuming rand() is uniformly distributed, not seeing any number that can't be generated or any that would be generated more or less often. With that said, ran this against dieharder 3.31.1; there were two "good" tests that came back weak [not failed] (as opposed to just using the output of rand(), which had many failures, presumably due to rand() not return 32 bits of randomness at a time, or with uchar x = rand() % 256 and outputting 8 bits at a time; only 1 suspect test came back weak (rest passed)
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If you want just to produce unsigned long long from value returned by rand() and do not care about the characteristics of the result consider the following function that must be compiler version and platform independent (because no "magic numbers" are used):

// this header has RAND_MAX value #include <stdlib.h> // and this header has ULLONG_MAX #include <limits.h> unsigned long long ullrand() // Produces pseudo-random numbers from 0 to ULLONG_MAX // by filling all bits of unsigned long long integer number // with bits of several "small" integer numbers generated by rand() { unsigned long long myrndnum = 0; // at the beginning just zero unsigned long long counter = ULLONG_MAX; // at the beginning we have all bits set as 1 // ... and while at least one bit is still set to 1 while(counter > 0) { myrndnum = (myrndnum * (RAND_MAX + 1)) + rand(); // fill some bits from rand() counter /= (RAND_MAX + 1); // decrease number of 1-bits in counter } // Return the result return myrndnum; } 

But if you want some sequence of random numbers with certain predetermined characteristics, you should look in some specific guides or math books. E.g. https://www.gnu.org/software/gsl/manual/html_node/Random-number-generator-algorithms.html

Comments

1

Assuming you want the outputs to be uniformly distributed, the standard code for this is to use std::uniform_int_distribution of the appropriate type:

#include <iostream> #include <random> int main() { auto gen = std::mt19937{std::random_device()()}; auto d = std::uniform_int_distribution<unsigned long long>{}; for (int i = 0; i < 10; ++i) { std::cout << d(gen) << '\n'; } } 

Note that while this can generate numbers smaller than 100,000, it won't happen very often - on average only once every 184,467,440,737,095 outputs if your unsigned long long is 64-bit.

Comments

0

You didn't ask for a specific OS and the answers here are really good, but on Linux (and probably on other OSes too) you can also read from a random device.

Example:

#include <stdio.h> #include <assert.h> #define RANDDEV "/dev/urandom" unsigned long long bigrand(void) { FILE *rdp; unsigned long long num; rdp = fopen(RANDDEV, "rb"); assert(rdp); assert(fread(&num, sizeof(num), 1, rdp) == 1); fclose(rdp); return num; } 

Written on mobile, may have bugs. :P

1 Comment

Maybe you should turn off buffering, or use read(), to avoid wasting random bytes?
0

You can also use the boost library (taken from link):

#include <ctime> // std::time #include <boost/random/mersenne_twister.hpp> #include <boost/random/linear_congruential.hpp> #include <boost/random/uniform_real.hpp> #include <boost/random/variate_generator.hpp> #include <boost/generator_iterator.hpp> int main() { long long my_min = 1; long long my_max = 1e5; boost::mt19937 generator(static_cast<unsigned int>(std::time(0))); boost::variate_generator<boost::mt19937&, boost::uniform_real<> > die_gen(generator, boost::uniform_real<> (my_min, my_max)); boost::generator_iterator<boost::variate_generator<boost::mt19937&, boost::uniform_real<> > > die(&die_gen); std::cout<<"Generated random numbers: \n"; for (int i=0; i <10 ; i++) { std::cout<< static_cast<long long>(*die++) << std::endl; } return 0; } 

Comments

-1

try this:

long N=1000000; long randNumber; for(long i=0;i<N;i++) randNumber=i+rand() 

2 Comments

While this code may answer the question, it would be better to include some context, explaining how it works and when to use it. Code-only answers are not useful in the long run.
This is a terrible method to generate random numbers. It's slow, badly distributed, and causes undefined behavior due to signed overflow.

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