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Home / How to Generate Random Numbers in C++ With a Range

How to Generate Random Numbers in C++ With a Range

6 Answers 6

Since nobody posted the modern C++ approach yet,

                #include <iostream> #include <random> int main() {     std::random_device rd; // obtain a random number from hardware     std::mt19937 gen(rd()); // seed the generator     std::uniform_int_distribution<> distr(25, 63); // define the range      for(int n=0; n<40; ++n)         std::cout << distr(gen) << ' '; // generate numbers }

answered Sep 26 '11 at 19:54

7

  • If you wonder "what the hack is mt19937 type?!" - A Mersenne Twister pseudo-random generator of 32-bit numbers with a state size of 19937 bits. It is a "wrapper" for the "mersenne_twister_engine" template (cplusplus.com/reference/random/mersenne_twister_engine) with pre-set params.

    Nov 12 '14 at 11:56

  • @Akiva I think, the last is correct. If it's absent from your distribution, you should upgrade your compiler to C++11-compliant one. It isn't a deprecated function, it's a new one. GCC 4.9+ should certainly have it.

    Jan 25 '15 at 0:41

  • Also worth noting that to do this same thing with floats/doubles, one can use std::uniform_real_distribution<>. Great answer!

    Jun 16 '15 at 20:10

  • Note the std::uniform_int_distribution<> is inclusive -- [25,63] in the above example.

    Sep 23 '16 at 4:20

  • @Akiva and @polkovnikov.ph: It's not about upgrading the compiler (GCC has had C++11 support for ages), but about switching to C++11 using the -std=c++11 compiler flag. Since C++11 is not backwards compatible, all standard-complying compilers (GCC, clang,...) require C++11 to be activated explicitly.

    Oct 27 '16 at 12:34

You can use the random functionality included within the additions to the standard library (TR1). Or you can use the same old technique that works in plain C:

                25 + ( std::rand() % ( 63 - 25 + 1 ) )

answered Sep 26 '11 at 19:21

4

  • Note that this answer does not generate equally-likely numbers! The below answer is a better approach.

    Mar 28 '14 at 7:23

  • @GeorgeHilliard why? i really does not know, how the algorithm for generating random numbers work

    Jul 29 '20 at 21:43

  • @milanHrabos an example will easily make that clear: std::rand() returns equally distributed integers from 0 to RAND_MAX (inclusively). Now, assuming you obtained integers in the range from 0 to RAND_MAX - 1 (inclusively) by using std::rand() % RAND_MAX, the chance of getting a 0 would now be doubled, since it will be the result when std::rand() returns either 0 or RAND_MAX. Any other number will only be obtained when directly returned by std::rand(). This will apply similarly, for every modulo that is not a divisor of RAND_MAX + 1.

    Aug 27 '20 at 14:29

  • @GeorgeHilliard "Below" is relative.

    Oct 22 at 7:06 

                int random(int min, int max) //range : [min, max] {    static bool first = true;    if (first)     {         srand( time(NULL) ); //seeding for the first time only!       first = false;    }    return min + rand() % (( max + 1 ) - min); }

answered Sep 26 '11 at 19:22

5

  • Hope you don't mind the edit, I'm pretty sure that was your intent.

    Sep 26 '11 at 19:45

  • For this to be inclusive of max you have to use max - min + 1

    Apr 25 '14 at 12:23

  • Does it include a range between negative to positive?

    Apr 21 '16 at 21:44

  • Use "return min + rand() % (max + 1 - min);" to obtain the max value itself! =D

    Oct 10 '16 at 17:25

  • rand() returns only numbers in band [0;RAND_MAX]. If (max - min) > RAND_MAX, you never got numbers in range (RAND_MAX; max - min). RAND_MAX is implementation details, but usually equals 7FFF.

    Oct 17 at 6:56 

                int range = max - min + 1; int num = rand() % range + min;

answered Sep 26 '11 at 19:21

7

  • Bad advice. This will not yield a uniform distribution. It is a fairly common approach though (but fundamentally flawed).

    Sep 26 '11 at 19:23

  • Why is this flawed? If rand() % range is relatively uniform, I would expect that adding a constant would preserve uniformity. Or are you saying that rand() itself isn't a great choice for a truly random number? That's true, but depending on what OP wants to do, rand() may be sufficient.

    Sep 26 '11 at 19:32

  • @Anson: The reason that Johannes Rudolph is complaining is that % does not work unless the value on the right is an exact divisor of RAND_MAX. If it is not then some values have a slight (very slight) smaller probability. example: RAND_MAX (32767) r = rand()%3; Probability of 0(10923/32767) 1(10922/32767) 2(10922/32767) Notice the probability of a 0 is slightly higher. As 'Benjamin Lindley' points out this is insignificant in most situations but when you do things like cryptography things like this become a problem. Thus a lot of people cry foul when this method is used.

    Sep 26 '11 at 20:25

  • Also there was a problem with the bottom few bits of rand() not being very random (thus if max-min is small you don't get good values (this may have been fixed)). The alternative being num = rand() * 1.0 / RAND_MAX * (max-min+1) + min;

    Sep 26 '11 at 20:29

  • @MartinYork your alternative is no better, it still suffers from the pigeonhole problem, it's just not as obvious why.

    May 8 at 18:37

Use the rand function:

http://www.cplusplus.com/reference/clibrary/cstdlib/rand/

Quote:

                A typical way to generate pseudo-random numbers in a determined range using rand is to use the modulo of the returned value by the range span and add the initial value of the range:  ( value % 100 ) is in the range 0 to 99 ( value % 100 + 1 ) is in the range 1 to 100 ( value % 30 + 1985 ) is in the range 1985 to 2014

answered Sep 26 '11 at 19:22

4

  • @Tux-D That was a code snippet copied straight from the cplusplus website so I guess I figured it was the standard way of doing it. I didn't realize it was incorrect or uncommon to do this way, thanks for the info. Because now I'm curious, what exactly is the problem with this method?

    Sep 27 '11 at 2:15

  • See Benjamin Lindley answer above and my associated comments for the pre-C++11 approach. But my comment above was wrong (I miss read your answer). So I will delete it. The website cplusplus.com is OK bit I (and good for quick reference now and then) but I would NOT use it as an authoritative source.

    Sep 27 '11 at 3:49

  • Thanks for the information. I read the information above and feel I have a better understanding. I would argue that while my answer isn't the best possible answer, it's not NOT useful. But that's your decision and I thank you again for enlightening me. :)

    Sep 27 '11 at 13:18 

                float RandomFloat(float min, float max) {     float r = (float)rand() / (float)RAND_MAX;     return min + r * (max - min); }

answered Sep 26 '11 at 19:22

4

  • That's not a uniform distribution. Try putting 7 balls into 3 buckets, such that each bucket has the same amount of balls. You're doing exactly the same.

    Jan 25 '15 at 0:43

  • I actually think its correct, its multiply a range by a number in [0,1]. Can you exmplain further @polkovnikov.ph ?

    Aug 27 '16 at 9:56

  • @LorenzoBelli Floats are not uniformly distributed over [0, 1], because they have limited precision. Real numbers are.

    Aug 27 '16 at 19:58

  • Tell me, what's the odds of this algorithm returning max?

    May 8 at 18:41

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How to Generate Random Numbers in C++ With a Range

Source: https://stackoverflow.com/questions/7560114/random-number-c-in-some-range/7560151