Jamie Keeling - 1 year ago 74

C Question

This is a follow on from a previously posted question:

How to generate a random number in C?

I wish to be able to generate a random number from within a particular range, such as 1 to 6 to mimic the sides of a dice.

How would I go about doing this?

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Answer Source

All the answers so far are mathematically wrong. Returning `rand() % N`

does not uniformly give a number in the range `[0, N)`

unless `N`

divides the length of the interval into which `rand()`

returns (i.e. is a power of 2). Furthermore, one has no idea whether the moduli of `rand()`

are independent: it's possible that they go `0, 1, 2, ...`

, which is uniform but not very random. The only assumption it seems reasonable to make is that `rand()`

puts out a Poisson distribution: any two nonoverlapping subintervals of the same size are equally likely and independent. For a finite set of values, this implies a uniform distribution and also ensures that the values of `rand()`

are nicely scattered.

This means that the only correct way of changing the range of `rand()`

is to divide it into boxes; for example, if `RAND_MAX == 11`

and you want a range of `1..6`

, you should assign `{0,1}`

to 1, `{2,3}`

to 2, and so on. These are disjoint, equally-sized intervals and thus are uniformly and independently distributed.

The suggestion to use floating-point division is mathematically plausible but suffers from rounding issues in principle. Perhaps `double`

is high-enough precision to make it work; perhaps not. I don't know and I don't want to have to figure it out; in any case, the answer is system-dependent.

The correct way is to use integer arithmetic. That is, you want something like the following:

```
#include <stdlib.h> // For random(), RAND_MAX
// Assumes 0 <= max <= RAND_MAX
// Returns in the closed interval [0, max]
long random_at_most(long max) {
unsigned long
// max <= RAND_MAX < ULONG_MAX, so this is okay.
num_bins = (unsigned long) max + 1,
num_rand = (unsigned long) RAND_MAX + 1,
bin_size = num_rand / num_bins,
defect = num_rand % num_bins;
long x;
do {
x = random();
}
// This is carefully written not to overflow
while (num_rand - defect <= (unsigned long)x);
// Truncated division is intentional
return x/bin_size;
}
```

The loop is necessary to get a perfectly uniform distribution. For example, if you are given random numbers from 0 to 2 and you want only ones from 0 to 1, you just keep pulling until you don't get a 2; it's not hard to check that this gives 0 or 1 with equal probability. This method is also described in the link that nos gave in their answer, though coded differently. I'm using `random()`

rather than `rand()`

as it has a better distribution (as noted by the man page for `rand()`

).

If you want to get random values outside the default range `[0, RAND_MAX]`

, then you have to do something tricky. Perhaps the most expedient is to define a function `random_extended()`

that pulls `n`

bits (using `random_at_most()`

) and returns in `[0, 2**n)`

, and then apply `random_at_most()`

with `random_extended()`

in place of `random()`

(and `2**n - 1`

in place of `RAND_MAX`

) to pull a random value less than `2**n`

, assuming you have a numerical type that can hold such a value. Finally, of course, you can get values in `[min, max]`

using `min + random_at_most(max - min)`

, including negative values.

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