array of non repeating numbers in C?

Question

I have an array of six random numbers from 1-49 so far but some repeat eg, 12 15 43 43 22 15 is there a way around this problem?

so far I have...

int* get_numbers()
{
        int Array[6], i;
        printf("\n\n Your numbers are : ");
        for (i = 0; i < 6; i++)
        {
               Array[i] = ((rand() % 49) + 1);
               printf("%d ",Array[i]);
        }
}

any feedback will be great, thanks

What is the problem you are trying to solve? Do you need 6 unique numbers? — , Mar 19 '14 at 18:08

score 2 · Answer 1 · edited May 23 '17 at 11:57

2

You could simply throw out duplicates.

Another option would be to create an array of numbers 1-49, shuffle them, and take the first 6.

edited May 23 '17 at 11:57

Community

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answered Mar 19 '14 at 18:08

Fred Larson

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Sergey Kalinichenko · Accepted Answer · 2014-03-19T18:13:41.397

You need to add a separate loop that goes from 0, inclusive, to i, exclusive, checking the candidate number against the numbers that have been added to the array before. If the check finds a duplicate, do not increment i, and try generating a random number again:

int i = 0;
while (i != 6) {
   int candidate = ((rand() % 49) + 1);
   int ok = 1; // ok will become 0 if a duplicate is found
   for (int j = 0 ; ok && j != i ; j++) {
       ok &= (candidate != Array[j]);
   }
   if (ok) {
       Array[i++] = candidate;
       printf("%d ", candidate);
   }
}

Demo on ideone.

score 0 · Answer 3 · answered Mar 19 '14 at 18:10

0

The rand function is not aware of its previous output so you can put a check before you save it in the array

answered Mar 19 '14 at 18:10

Am_Sri

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score 0 · Answer 4 · answered Mar 19 '14 at 18:11

0

If you get the first random number as x, find the second random number as the random of random(1 to x-1) and random(x+1 to n). Continue in this way.

answered Mar 19 '14 at 18:11

HelloWorld123456789

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array of non repeating numbers in C?

4 Answers4