Question

int main() 
{ 
  srand((unsigned)time(0)); 
  int random_integer; 
  int lowest=0, highest=10; 
  int range=(highest-lowest)+1; 
  for(int index=0; index<20; index++){ 
    random_integer = (rand() % range) + lowest/(RAND_MAX + 1.0);
    cout << random_integer << endl; 
  } 
} 

I am getting the output from 0 to 10, 11 numbers, but I don't want to get number 10, just numbers 0 to 9, that means 10 random numbers, what should I do?

Was it helpful?

Solution

Modulo operation x % c; returns the remainder of division of number x by c. If you do x % 10 then there are 10 possible return values: 0 1 2 3 4 5 6 7 8 9.

Note that generating random numbers by using rand() with % produces skewed results and those numbers are not uniformly distributed. Here's the simple C-style function that generates random number from the interval from min to max, inclusive:

int irand(int min, int max) {
    return ((double)rand() / ((double)RAND_MAX + 1.0)) * (max - min + 1) + min;
}

Note, that numbers generated by this functions are uniformly distributed:

int occurences[8] = {0};

srand(time(0));
for (int i = 0; i < 100000; ++i)
    ++occurences[irand(1,7)];

for (int i = 1; i <= 7; ++i)
    cout << occurences[i] << ' ';

output: 14253 14481 14210 14029 14289 14503 14235

Also have a look at:
Generate a random number within range?
Generate random numbers uniformly over an entire range
What is the best way to generate random numbers in C++?

OTHER TIPS

It appears that you need to fully understand, then modify this line

random_integer = (rand() % range) + lowest/(RAND_MAX + 1.0);

rand() % 10 normally does the trick for me.. (unless I'm missing something)

int main() 
{ 
  srand((unsigned)time(0)); 
  int random_integer; 
  int lowest=0, highest=9; 
  int range=(highest-lowest)+1; 
  for(int index=0; index<20; index++){ 
    random_integer = (rand() % range) + lowest;
    cout << random_integer << endl; 
  } 
} 

this type of questions have been answered n number of times.

have a look at this

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