I need to generate random integers inside a range [a, b] in C. I used the normal implementation that you will see everywhere on the internet and wrote the following:

 * randRange Generates random integers in range
 * @param lower Lower bound
 * @param upper Upper bound
 * @return random integer between lower and upper (inclusive)
int randRange(int lower, int upper) {
    return rand() % (++upper - lower) + lower;

Now, this works just fine, but the distribution over the range really isn't good, and it's starting to cause secondary algorithms I have to behave badly. How can I improve that function so that it is:

  1. Not biased
  2. Doesn't rely on anything not on the standard library
  3. Is suitable for a Monte Carlo algorithm
  4. Fast

1 Answer 1


There are many good reasons to not use modulo for random numbers like this:

LCGs have poor entropy in lower bits

It's well known that the lower bits on Linear Congruential Generators (LCG) have poor entropy. And although not required to be, rand() is typically an LCG.

This means that using:

rand()%range + min 

has even worse entropy as it is only using the lower bits due to the modulo operator.

Using modulo skews the distribution

As if low entropy wasn't enough, using modulo in this way also skews the distribution of numbers.

To see this, imagine RAND_MAX=3 and you do rand()%3 then the possible outcomes are:

0%3 = 0
1%3 = 1
2%3 = 2
3%3 = 0

Zero is twice as likely as any other outcome. This is clearly not uniform! Even for larger RAND_MAX and range values the problem persists. There are some special cases where it works but in general it doesn't.

So how should you do it?

Unfortunately though you see the approach with modulo shown everywhere on the internet because it's easier to understand. Or because people don't know better. Whatever the reason using modulo is not the way to limit the range of a random number generator.

There are a few different ways:

Use all the bits - Floating point rescale

As rand() returns a some what uniform random number in the range [0,RAND_MAX] and we want all numbers to be used; So we can re-scale this range to [0,1[ and multiply it by our desired range and quantize it. This requires a bit of care so that we don't introduce additional bias around the edges of the range:

// Use whatever c equivalent of next_after get one ULP larger than
// RAND_MAX. This transforms the range from to [0,1[ with minimal
// skew in the distribution. 
double normalized = rand() / std::next_after(double(RAND_MAX), DOUBLE_MAX);

int rnd = min + int(normalized * (max - min + 1));

This method while better than the modulo in the aspect that it has higher entropy as it uses the high bits of the rand() call still suffers some bias problems. Consider again if RAND_MAX=3, min=0 and max=2;

rand()=0 -> norm = 0                       rnd = 0
rand()=1 -> norm = 0.333333333333333332    rnd = 0
rand()=2 -> norm = 0.666666666666666665    rnd = 1
rand()=3 -> norm = 0.9999999999999999998   rnd = 2

the good news is that as RAND_MAX grows larger, the bias approaches (but never reaches) zero as 1/x (which is also true when using the modulo approach but exaggerated by the bad entropy in the lower bits).

For example for RAND_MAX=3000 you will have 1001 trials that result in rnd=0 and 1000 trials each that result in rnd=1 or rnd=2.

Depending on your requirements, this may or may not be enough.

Use all the bits - Integer only w/ re-rolls

So we want to use the whole range and use integers only, if we split the range of random numbers from rand() into equally large "chunks" and depending on which chunk we "land in" we determine the outcome. In the general case there will be some slack at the end of the range as the chunk size doesn't evenly divide the range, if we roll a value in the slack we just roll again:

int range = max - min + 1;

// Largest value that when multiplied by "range"
// is less than or equal to RAND_MAX
int chunkSize = (RAND_MAX + 1) / range; 
int endOfLastChunk = chunkSize * range;

int r = rand();
while(r >= endOfLastChunk){
    r = rand();
return min + r / chunkSize;

We have to be careful about the inequalities and off by ones here as any error adds a bias. Lets check:

min = 0;
max = 2;

range = 3;
chunkSize = (5 + 1)/3 = 2;
endOfLastChunk = 2 * 3 = 6;

r = 0 -> return       0;
r = 1 -> return 1/2 = 0;
r = 2 -> return 2/2 = 1;
r = 3 -> return 3/2 = 1;
r = 4 -> return 4/4 = 2;
r = 5 -> return 5/4 = 2;

Yay! No bias! Lets try again with RAND_MAX = 4;

chunkSize = (4 + 1)/3 = 1;
endOfLastChunk = 1 * 3 = 3;

r = 0 -> return       0;
r = 1 -> return 1/1 = 1;
r = 2 -> return 2/1 = 2;
r = 3 -> reroll
r = 4 -> reroll

As you can see no bias here.

The things to take note of: The reroll takes care of the bias; and using the largest possible chunkSize and thus using the largest part possible of the output range of rand() avoids the issue with the low bits being of poor quality (unless you have a very large range but then you're SOL anyway).

The probability of a re-roll is:

(RAND_MAX + 1 - endOfLastChunk) / (RAND_MAX + 1)

note that: (RAND_MAX + 1 - endOfLastChunk) < range otherwise chunkSize would be bigger. So we have:

 range / (RAND_MAX + 1)

chance of a re-roll.

In closing

That all being said and done, linear congruential generators are generally poor choices for any kind of serious pseudo random numbers overall, you should look into a better generator like for example the widely used Mersenne Twister.

  • \$\begingroup\$ Note that your version which divides by double(RAND_MAX) will also have a bias towards certain numbers. In your example with RAND_MAX = 3, min = 0, max = 2, your version will generate the value 0 twice just like the modulo version. Actually, it is impossible to fairly map RAND_MAX input values to range output values when RAND_MAX is not evenly divisible by range. But I do agree that the lower bits of LCGs are less random than the upper bits, so your version will generate better randomness. \$\endgroup\$
    – JS1
    Apr 3, 2017 at 22:24
  • \$\begingroup\$ @JS1 rand()/double(RAND_MAX) is equally uniform in [0, 1] as rand() is uniform in [0, RAND_MAX], i.e. as good as it's going to get with an LCG. The problem is how you scale and then quantise your range which may or may not create a bias which I hinted at with "check the bounds" and cautions further down in my answer. I'll give a more thorough edit tomorrow when I'm awake and by a computer :) \$\endgroup\$
    – Emily L.
    Apr 3, 2017 at 22:55
  • \$\begingroup\$ I was referring to (rand()/double(RAND_MAX))*(max-min))+min, which will exhibit bias towards certain numbers in the range. But the main point was that it isn't possible to scale/quantize from RAND_MAX to range fairly, without the ability to throw away some results and getting one or more additional random numbers. \$\endgroup\$
    – JS1
    Apr 4, 2017 at 0:03
  • \$\begingroup\$ @JS1 Now, when it's not 3am, I see your point and I have updated the answer. \$\endgroup\$
    – Emily L.
    Apr 4, 2017 at 5:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.