Solution for Exercise 11-18

ex1118

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int r,a,b;

    puts("100 Random Numbers");
    for(a=0;a<20;a++)
    {
        for(b=0;b<5;b++)
        {
            r=rand();
            r%=21;
            printf("%d\t",r);
        }
        putchar('\n');
    }
    return(0);
}

Output

100 Random Numbers
7   7   2   20  13
20  9   8   8   1
6   2   0   20  0
5   16  4   4   10
18  19  15  14  1
3   6   9   14  16
4   17  14  13  8
12  7   1   0   18
13  14  7   0   2
5   11  4   6   16
2   5   7   1   7
17  9   14  6   7
19  7   12  17  8
1   0   0   9   10
7   13  19  13  0
1   14  12  20  5
6   10  0   7   8
17  18  17   3  3
18  8   11  16  12
20  13  12  11  3

Notes

* The zero is possible because r%=21 generates a value of zero when the random number coughed up by the rand() function is divisible by 21.

* You could also combine Lines 13 and 14 to read:

r=rand() % 21;

* Or you could set r%21 as the printf() statement's argument instead of just r. In this case, you could remove Line 14 from the code. And you could eliminate int variable r from the code if you set rand()%21 as the second argument in the printf() statement.

* Most of the random number generating routine you use incorporate the modulus operator to curb the values, just as shown above.

Exercise ex1119 →