Count the numbers of a random generator in C

user_16673050 0
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My code generates a random number between 1 and 6 for 1000 times and I need to count how many times each number appeared in percentage. My problem is that it is, sometimes, generating negative percentages and it is surpassing 100%. I also must use a switch.

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

int main(void) {
    int i, n, max = 6;
    int num1, num2, num3, num4, num5, num6;
    float total1, total2, total3, total4, total5, total6;

    srand(time(NULL));
    for (i = 0; i < 1000; i++) {
        n = rand() % max + 1; 
        switch (n) {
          case 1:
            num1 += 1;
          case 2:
            num2 += 1;
          case 3:
            num3 += 1;
          case 4:
            num4 += 1;
          case 5:
            num5 += 1;
          case 6:
            num6 += 1;
        }
    }
    total1 = num1 / 100;
    printf("Times 1 appeared: %f\n", total1);
    total2 = num2 / 100;
    printf("Times 2 appeared: %f\n", total2);
    total3 = num3 / 100;
    printf("Times 3 appeared: %f\n", total3);
    total4 = num4 / 100;
    printf("Times 4 appeared: %f\n", total4);
    total5 = num5 / 100;
    printf("Times 5 appeared: %f\n", total5);
    total6 = num6 / 100;
    printf("Times 6 appeared: %f\n", total6);
}
c
2 Answers

There are several problems:

  • Several values were not initialized to 0. This needs to happen explicitly.
  • The switch cases had fallthrough. Landing on, e.g., case 1 meant that the execution continued through the other cases below it.
  • The totals were computed using integer division instead of floating point division, leading to a lack of precision.
  • The decimal point for the totals was in the wrong place.

The improved code:

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

int main(void) {
    int i, n, max = 6;
    int num1 = 0, num2 = 0, num3 = 0, num4 = 0, num5 = 0, num6 = 0;    
    float total1 = 0.0F, total2 = 0.0F, total3 = 0.0F, total4 = 0.0F, total5 = 0.0F, total6 = 0.0F;

    srand(time(NULL));
    for (i = 0; i < 1000; i++) {
        n = rand() % max + 1; 
        switch (n)
        {
        case 1:
            num1 += 1;
            break;
        case 2:
            num2 += 1;
            break;
        case 3:
            num3 += 1;
            break;
        case 4:
            num4 += 1;
            break;
        case 5:
            num5 += 1;
            break;
        case 6:
            num6 += 1;
            break;
        }
    }
    total1 = num1/10.0F;
    printf("Times 1 appeared: %f\n", total1);
    total2 = num2/10.0F;
    printf("Times 2 appeared: %f\n", total2);
    total3 = num3/10.0F;
    printf("Times 3 appeared: %f\n", total3);
    total4 = num4/10.0F;
    printf("Times 4 appeared: %f\n", total4);
    total5 = num5/10.0F;
    printf("Times 5 appeared: %f\n", total5);
    total6 = num6/10.0F;
    printf("Times 6 appeared: %f\n", total6);
}

The code could be much more compact while achieving the same results. In C17:

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

#define NUMBER_OF_BINS 6

int main(void)
{
    unsigned int number_of_iterations = 1000U;
    unsigned counters[NUMBER_OF_BINS] = {0U};

    srand(time(NULL));
    for (unsigned i = 0; i < number_of_iterations; ++i)
        ++counters[rand() % NUMBER_OF_BINS];

    for (size_t idx = 0; idx < NUMBER_OF_BINS; ++idx)
        printf("Times %zu appeared: %f\n", idx + 1, counters[idx] * 100.0F / number_of_iterations);
}
user_166322810

A better approach would be to use an array of ints to store the frequency of each number, if you could avoid the use of switch of course ,since you are dealing with a small range , 1 - 6, it can save you a lot of time. Should look like this,

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

#define MAX 6

int main(void)
{
    int i, n;
    int num[MAX] = {0}; /* initialize to 0 */

    srand(time(NULL));

    for (i = 0; i < 1000; i++)
    {
        n = rand() % MAX + 1;

        num[n - 1]++; /* since arrays start from 0 , decrement by one to avoid overflow */
    }

    for (int i = 0; i < ΜΑΧ; i++)
    {
        printf("number %d, appeared :  %.1f%%\n", i + 1, ((float)num[i] * 100.0f / 1000.0f));
    }

    return 0;
}
user_126907180
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