# c – Why does rand() + rand() produce negative numbers?

## The Question :

306 people think this question is useful

I observed that rand() library function when it is called just once within a loop, it almost always produces positive numbers.

for (i = 0; i < 100; i++) {
printf("%d\n", rand());
}



But when I add two rand() calls, the numbers generated now have more negative numbers.

for (i = 0; i < 100; i++) {
printf("%d = %d\n", rand(), (rand() + rand()));
}



Can someone explain why I am seeing negative numbers in the second case?

PS: I initialize the seed before the loop as srand(time(NULL)).

• rand() can’t be negative…
• rand() + rand() can owerflow
• What is RAND_MAX for your compiler? You can usually find it in stdlib.h. (Funny: checking man 3 rand, it bears the one-line description “bad random number generator”.)
• do what every sane programmer would do abs(rand()+rand()). I’d rather have a positive UB than a negative one! 😉
• @hexa: that is no sotution for the UB, as the occurs for the addition already. You cannot make UB become defined behaviour. A sane progrtammer would avoid UB like hell.

546 people think this answer is useful

rand() is defined to return an integer between 0 and RAND_MAX.

rand() + rand()



could overflow. What you observe is likely a result of undefined behaviour caused by integer overflow.

91 people think this answer is useful

The problem is the addition. rand() returns an int value of 0...RAND_MAX. So, if you add two of them, you will get up to RAND_MAX * 2. If that exceeds INT_MAX, the result of the addition overflows the valid range an int can hold. Overflow of signed values is undefined behaviour and may lead to your keyboard talking to you in foreign tongues.

As there is no gain here in adding two random results, the simple idea is to just not do it. Alternatively you can cast each result to unsigned int before the addition if that can hold the sum. Or use a larger type. Note that long is not necessarily wider than int, the same applies to long long if int is at least 64 bits!

Conclusion: Just avoid the addition. It does not provide more “randomness”. If you need more bits, you might concatenate the values sum = a + b * (RAND_MAX + 1), but that also likely requires a larger data type than int.

As your stated reason is to avoid a zero-result: That cannot be avoided by adding the results of two rand() calls, as both can be zero. Instead, you can just increment. If RAND_MAX == INT_MAX, this cannot be done in int. However, (unsigned int)rand() + 1 will do very, very likely. Likely (not definitively), because it does require UINT_MAX > INT_MAX, which is true on all implementations I’m aware of (which covers quite some embedded architectures, DSPs and all desktop, mobile and server platforms of the past 30 years).

Warning:

Although already sprinkled in comments here, please note that adding two random values does not get a uniform distribution, but a triangular distribution like rolling two dice: to get 12 (two dice) both dice have to show 6. for 11 there are already two possible variants: 6 + 5 or 5 + 6, etc.

Also note that the results rand() generates are not independent of each other, as they are generated by a pseudorandom number generator. Note also that the standard does not specify the quality or uniform distribution of the calculated values.

36 people think this answer is useful

the reason i was adding was to avoid ‘0’ as the random number in my code. rand()+rand() was the quick dirty solution which readily came to my mind.

The problem was to avoid 0. There are (at least) two problems with the proposed solution. One is, as the other answers indicate, that rand()+rand() can invoke undefined behavior. Best advice is to never invoke undefined behavior. Another issue is there’s no guarantee that rand() won’t produce 0 twice in a row.

The following rejects zero, avoids undefined behavior, and in the vast majority of cases will be faster than two calls to rand():

int rnum;
for (rnum = rand(); rnum == 0; rnum = rand()) {}
// or do rnum = rand(); while (rnum == 0);



3 people think this answer is useful

Basically rand() produce numbers between 0 and RAND_MAX, and 2 RAND_MAX > INT_MAX in your case.

You can modulus with the max value of your data-type to prevent overflow. This ofcourse will disrupt the distribution of the random numbers, but rand is just a way to get quick random numbers.

#include <stdio.h>
#include <limits.h>

int main(void)
{
int i=0;

for (i=0; i<100; i++)
printf(" %d : %d \n", rand(), ((rand() % (INT_MAX/2))+(rand() % (INT_MAX/2))));

for (i=0; i<100; i++)
printf(" %d : %ld \n", rand(), ((rand() % (LONG_MAX/2))+(rand() % (LONG_MAX/2))));

return 0;
}



2 people think this answer is useful

May be you could try rather a tricky approach by ensuring that the value returned by sum of 2 rand() never exceeds the value of RAND_MAX. A possible approach could be sum = rand()/2 + rand()/2; This would ensure that for a 16 bit compiler with RAND_MAX value of 32767 even if both rand happens to return 32767, even then (32767/2 = 16383) 16383+16383 = 32766, thus would not result in negative sum.

1 people think this answer is useful

the reason i was adding was to avoid ‘0’ as the random number in my code. rand()+rand() was the quick dirty solution which readily came to my mind.

A simple solution (okay, call it a “Hack”) which never produces a zero result and will never overflow is:

x=(rand()/2)+1    // using divide  -or-
x=(rand()>>1)+1   // using shift which may be faster
// compiler optimization may use shift in both cases



This will limit your maximum value, but if you don’t care about that, then this should work fine for you.

1 people think this answer is useful

To avoid 0, try this:

int rnumb = rand()%(INT_MAX-1)+1;



You need to include limits.h.

0 people think this answer is useful

thx. the reason i was adding was to avoid ‘0’ as the random number in my code. rand()+rand() was the quick dirty solution which readily came to my mind

It sounds like an XY problem to me, in which in order to not get a 0 from rand(), you call rand() two times, doing the program slower, with a new setback and the possibility of getting a 0 is still there.

Another solution is using uniform_int_distribution, which creates a random and uniformly distributed number in the defined interval:

#include <random>
#include <array>
#include <iostream>

int main()
{
const int MAX_VALUE=50;
const int MIN_VALUE=1;
std::random_device rd;
std::mt19937 gen(rd());
std::uniform_int_distribution<> distrib(MIN_VALUE, MAX_VALUE);
std::array<int,MAX_VALUE-MIN_VALUE> weight={0};

for(int i=0; i<50000; i++) {
weight[distrib(gen)-MIN_VALUE]++;
}

for(int i=0;i<(int)weight.size();i++) {
std::cout << "value: " << MIN_VALUE+i << " times: " << weight[i] << std::endl;
}
}