Generate random number bigger than 32767

Question

I am trying to assign to a variable a number between 0 and 65535 exactly, but i can't because RAND_MAX is set to 32767. How can I generate a random number bigger than 32767 and at the same time with the possibility that the number generated is between 0 and 65535?

Following up on @Brian, please read [this excellent article on the use of `` tools and why `rand()` is generally bad](http://cpp.indi.frih.net/blog/2014/12/the-bell-has-tolled-for-rand/). — zopieux, Mar 07 '15 at 00:45

Milack27 · Answer 1 · 2015-03-07T00:53:08.753

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I would try something like this:

int msb = (rand() % 2) << 15;
int random = msb | rand();

That is, I generate two random numbers. The first one (msb) will set the Most Significant Bit of the second (random).

edited Mar 07 '15 at 00:53

answered Mar 07 '15 at 00:42

Milack27

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if int is 32 bits then msb could become much larger than 65535 after shifting. may be you mean short msb = rand() << 15. – engf-010 Mar 07 '15 at 00:49
You're right. I would have to mask the 16 most significant bits to have the desired effect. – Milack27 Mar 07 '15 at 00:52
Thanks. Two questions. What is << 15 doing?, and in the second line what is the | operator doing there? – Lovenkrands Mar 07 '15 at 02:32
The standard rand() function generates a random number of 15 bits, going from 0 to 32767, as you've mentioned. To extend this range to 65535, we have to add one more bit. So, in the first line, I'm generating a random number between 0 and 1, that is, rand() % 2. The << 15 is shifting this result 15 bits to the left, turning msb into the most significant bit. The | operator is a bitwise OR, and its function in my code is to put the extra bit in the new random number generated. If I had used the + operator, the result would be the same (in this case, not always). – Milack27 Mar 07 '15 at 02:50
I understand. But, going deep, why is a | operator doing the function of the + operator; why it does work successfully? – Lovenkrands Mar 07 '15 at 02:59
Because in this case, msb is an unknown bit followed by 15 zeros. On the other hand, rand() is a zero bit followed by 15 unknown bits. Like this, in binary: msb = x000000000000000; rand() = 0xxxxxxxxxxxxxxx. So, msb + rand() = msb | rand() = xxxxxxxxxxxxxxxx. Consider x as an unknown bit. When doing an OR operation between a zero bit and x, the result is the same x. Because msb and rand() are "complementary", the bitwise OR results the same as a simple addition. – Milack27 Mar 07 '15 at 03:07

score 6 · Accepted Answer · answered Mar 07 '15 at 00:44

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If your compiler supports C++11 you can do something like this:

#include <iostream>
#include <random>

int main()
{
    std::random_device rd;
    std::default_random_engine eng {rd()};
    std::uniform_int_distribution<> dist(0, 65535);

    for(unsigned i = 0; i < 10; ++i)
        std::cout << dist(eng) << '\n';
}

Sample output:

answered Mar 07 '15 at 00:44

Galik

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Can you modify the syntaxis without the std:: ? (when "using namespace std;") – Lovenkrands Mar 07 '15 at 00:52
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@user3273960 I would never recommend doing that: https://stackoverflow.com/questions/1452721/why-is-using-namespace-std-considered-bad-practice but it would definitely work. – Galik Mar 07 '15 at 00:56
Yes, it's not because i want to but because i need to do it with namespace. Question: With namespace: uniform_int_distribution<> dist(0, 65535); stays like this? – Lovenkrands Mar 07 '15 at 02:45
@user3273960 yes, try it and see, just remove the `std::` qualifier. – Galik Mar 07 '15 at 04:12
@Galik, is the same thing achievable in C ? – JenilDave May 17 '20 at 06:55
@JenilDave Yes, you should be able to find some library code that does it. Otherwise @Milack27's answer will work in `C`. – Galik May 17 '20 at 07:34
@Galik, thanks for your help, but i figured out my way, i tried this : ```int start=50000; int stop =60000; srand(time(0)); int lsb = rand()%(stop-start); lsb = start+lsb;``` What do you think about this? – JenilDave May 17 '20 at 07:42

Generate random number bigger than 32767

2 Answers2