Using rand( ) to generate a random sequence of integers in C

Question

I am trying to implement a function int choose_N(void) that generates and returns a random list of integers N (up to four integers). I am not sure if I am using rand( ) in the correct way, but this is what I have so far:

int choose_N(void)
{
  int x, integer[4];
  srand ((int) time(NULL));
  for (x=0; x<4; x++)
   integer[x]=rand();
  return integer;
}

Would looping it like this work? Is time initialized in the correct way?

score 4 · Answer 1 · answered Mar 03 '14 at 01:36

Two problems: most importantly, the array integer is local, and will go away when you return. If you want to return an array of things in C, you have to make the caller pass it in (see below). Secondly, srand() should only be called once per program, so it doesn't belong inside a function like this--it should be in main or some similar initialization function, otherwise multiple calls to this function will be correlated (at worst, they might even be identical). So here's a better approach:

void choose_n(int *out, int count) {
    for (int i = 0; i < count; i += 1) {
        out[i] = rand();
    }
}

int main(int argc, char *argv[]) {
    int results[4];

    srand(time(NULL));
    choose_n(results, 4);
}

score 0 · Answer 2 · answered Mar 03 '14 at 01:29

Your loop is correct but your return statement will not work. The array integer is local to the function and will not be available outside. In either case, your function is supposed to return int and not an array.

And if you want to limit the random number to a range, you are better off using the mod operator such as rand() % range.

score 0 · Answer 3 · edited May 23 '17 at 10:29

0

You can't return a pointer to a local variable from a function. See this post for the reason why? Can a local variable's memory be accessed outside its scope?

You have to allocate the array on heap.

Change it to:

int choose_N(void)
{
  int x;
  int *integer = (int*)malloc(sizeof(int)*4);
  srand ((int) time(NULL));
  for (x=0; x<4; x++)
   integer[x]=rand();
  return integer;
}

Don't forget to free() the memory when you are done. In case you are wondering about malloc & free, you can read here:http://www.codingunit.com/c-tutorial-the-functions-malloc-and-free

edited May 23 '17 at 10:29

Community

1
1

answered Mar 03 '14 at 01:32

brokenfoot

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Note that this leaves the caller responsible for `free`ing the return value later on, something that user3303851 may not know :) – hobbs Mar 03 '14 at 01:33
Added a link on how to use malloc and free :) – brokenfoot Mar 03 '14 at 01:40

score 0 · Answer 4 · answered Mar 03 '14 at 01:34

0

You could define your choose_N() like this

void choose_N(int integer[4])
{
    int x;
    srand ((int) time(NULL));
    for (x=0; x<4; x++)
        integer[x]=rand();
}

and use it likes this

int integer[4];
choose_N(integer);

answered Mar 03 '14 at 01:34

Lee Duhem

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Any 2 calls in the same second would produce the same result. – chux - Reinstate Monica Mar 03 '14 at 02:39
@chux It is, and that is the expected behavior. – Lee Duhem Mar 03 '14 at 02:43

Piotr Zurkowski · Answer 5 · 2014-03-03T07:55:23.167

0

About the random part: If you really want to feel the randomness of c rand() seeded by time(0), you should run rand() like 3 times before you use it for generating actual numbers. To justify that, try running this code few times one after another while printing the numbers. You will notice that first one will be almost the same. My out:

1023384829 832292180 1773676349 957893636 
1023401636 1114767429 1248842775 1942837294 
1023418443 1397242678 724009201 780297305

So your srand should look like this:

  srand ((int) time(NULL));
  rand(); rand(); rand();

edited Mar 03 '14 at 07:55

answered Mar 03 '14 at 01:38

Piotr Zurkowski

186
9

Interesting. I don't see that behavior at all. – Fred Larson Mar 03 '14 at 01:52
This could actually be possible, as OSX uses different (following C90 standard) rand() implementation to linux. There is a guy mentioning it here http://stackoverflow.com/a/462350/2160680 – Piotr Zurkowski Mar 03 '14 at 01:58
1

This is just wrong. You should only call `srand()` once per program. Particularly if you're seeding by `time()`: If you do that in a short loop, you're likely to reuse the same "random" sequence, and discarding the first three generated values won't change that. There is a little value in discarding a few random numbers after calling srand() at the beginning of a program, if you expect the program to be repeated relatively often, but less often than once a second. `/dev/random` is a much better seed source, but you should still only seed once per executable. – rici Mar 03 '14 at 02:05
Yeah, definitely. I just wanted to mention the time(0) seed problem with some implementations. I didn't want to repeat Lee Daniel Crocker answer about moving srand out of the function. The results I posted wasn't obtained by calling function multiple times, but running the executable three times, so it always called srand(time(0)) once. – Piotr Zurkowski Mar 03 '14 at 02:10
"randomness of c++ rand()" - Is this not a C question? – chux - Reinstate Monica Mar 03 '14 at 02:29

Using rand( ) to generate a random sequence of integers in C

5 Answers5