Random number c++ in some range

Question

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Generate Random numbers uniformly over entire range

I want to generate the random number in c++ with in some range let say i want to have number between 25 and 63.

How can I have that?

this works perfectly even with loops: output = min + (rand() % (int)(max - min + 1)) — Hatim, Nov 25 '16 at 07:56

score 357 · Answer 1 · edited Jun 06 '20 at 20:24

357

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
}

edited Jun 06 '20 at 20:24

Dmitri Nesteruk

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answered Sep 26 '11 at 19:54

Cubbi

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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 (http://www.cplusplus.com/reference/random/mersenne_twister_engine/) with pre-set params. – jave.web Nov 12 '14 at 11:56
I made a little I/O console app that asks for info and then dynamicly outputs the random numbers acordingly - http://ideone.com/wk239Q *BTW: modern C++ approach - +♥ ! * – jave.web Nov 12 '14 at 12:11
Downvote for the moment, as `std::random_device` returns the error `/home/akiva/Programming/Tutorial/tutorial/main.cpp:21: error: 'random_device' is not a member of 'std'` in my ubuntu 14.10 setup with qtcreator. Not sure if this a deprecated function, that it is not included in Linux, or if I am simply incompetent. – Anon Dec 31 '14 at 12:07
7

@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. – polkovnikov.ph Jan 25 '15 at 00:41
10

Also worth noting that to do this same thing with floats/doubles, one can use std::uniform_real_distribution<>. Great answer! – David K. Jun 16 '15 at 20:10
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Note the `std::uniform_int_distribution<>` is inclusive -- `[25,63]` in the above example. – BoltzmannBrain Sep 23 '16 at 04:20
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@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. – mastov Oct 27 '16 at 12:34
could anybody explain how to put this stuff into class members without template overblowing and simply reuse the number generation on any place, like a loop or a callback function? [Here](https://cppbyexample.com/random_int.html) is a ref but only with globals – woodz May 03 '22 at 07:54
Thanks though it is incredible the hops one has to jump just to get a random integer in C++ in this day and age – gatopeich Oct 19 '22 at 11:44
@gatopeich that's why there's a proposal for https://en.cppreference.com/w/cpp/experimental/randint – Cubbi Oct 19 '22 at 23:06

score 50 · Accepted Answer · answered Sep 26 '11 at 19:21

50

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

K-ballo

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Note that this answer *does not generate equally-likely numbers!* The below answer is a better approach. – George Hilliard Mar 28 '14 at 07:23
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@GeorgeHilliard why? i really does not know, how the algorithm for generating random numbers work – milanHrabos Jul 29 '20 at 21:43
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@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`. – Reizo Aug 27 '20 at 14:29
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@GeorgeHilliard "Below" is relative. – stackprotector Oct 22 '21 at 07:06

score 31 · Answer 3 · answered Sep 26 '11 at 19:21

31

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

answered Sep 26 '11 at 19:21

Benjamin Lindley

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Bad advice. This will not yield a uniform distribution. It is a fairly common approach though (but fundamentally flawed). – Johannes Rudolph 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. – Anson Sep 26 '11 at 19:32
@Johannes: I think [this is uniform enough](http://ideone.com/y6m8o) for most peoples purposes unless they specifically mention perfect uniformity as a requirement. – Benjamin Lindley Sep 26 '11 at 19:40
adding a constant will not preserve uniformity when range is large. However, for most applications this is uniform enough. – Mooing Duck Sep 26 '11 at 19:42
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@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. – Martin York Sep 26 '11 at 20:25
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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;` – Martin York Sep 26 '11 at 20:29
1

@MartinYork your alternative is no better, it still suffers from the pigeonhole problem, it's just not as obvious why. – Mark Ransom May 08 '21 at 18:37

score 30 · Answer 4 · edited Sep 29 '20 at 20:25

30

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);
}

edited Sep 29 '20 at 20:25

Pavan Manjunath

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answered Sep 26 '11 at 19:22

Nawaz

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Hope you don't mind the edit, I'm pretty sure that was your intent. – Benjamin Lindley Sep 26 '11 at 19:45
3

For this to be inclusive of max you have to use max - min + 1 – uchamp Apr 25 '14 at 12:23
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Does it include a range between negative to positive? – Khalil Khalaf Apr 21 '16 at 21:44
1

Use "return min + rand() % (max + 1 - min);" to obtain the max value itself! =D – Eduardo Lucio 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. – Andrew Oct 17 '21 at 06:56

score 10 · Answer 5 · answered Sep 26 '11 at 19:22

10

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

Kiley Naro

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@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? – Kiley Naro Sep 27 '11 at 02:15
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. – Martin York Sep 27 '11 at 03:49
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. :) – Kiley Naro Sep 27 '11 at 13:18
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See: http://www.azillionmonkeys.com/qed/random.html – Martin York Sep 27 '11 at 15:32

score 3 · Answer 6 · answered Sep 26 '11 at 19:22

3

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

Yurii Hohan

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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. – polkovnikov.ph Jan 25 '15 at 00:43
I actually think its correct, its multiply a range by a number in [0,1]. Can you exmplain further @polkovnikov.ph ? – Lorenzo Belli Aug 27 '16 at 09:56
1

@LorenzoBelli Floats are not uniformly distributed over [0, 1], because they have limited precision. Real numbers are. – polkovnikov.ph Aug 27 '16 at 19:58
Tell me, what's the odds of this algorithm returning `max`? – Mark Ransom May 08 '21 at 18:41

Random number c++ in some range

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