Fill an Array With Random Numbers

Fill an array with random numbers

You need to add logic to assign random values to double[] array using randomFill method.

Change

 public static double[] list(){
    anArray = new double[10];   
    return anArray;
 }

To

 public static double[] list() {
    anArray = new double[10];
    for(int i=0;i<anArray.length;i++)
    {
        anArray[i] = randomFill();
    }
    return anArray;
}

Then you can call methods, including list() and print() in main method to generate random double values and print the double[] array in console.

 public static void main(String args[]) {

list();
print();
 }

One result is as follows:

-2.89783865E8 
1.605018025E9 
-1.55668528E9 
-1.589135498E9 
-6.33159518E8 
-1.038278095E9 
-4.2632203E8 
1.310182951E9 
1.350639892E9 
6.7543543E7 

Fill array with random numbers within a specified range (C++)

You are using exact same integer in each initialization. for loop should be like that

    for(int i = 0; i < size; i++){

        arr[i] =  (rand() % 10);
     }

How to fill an array with random numbers from 0 to 99 using the class Math?

It should be like

 ar1[i] = (int)(Math.random() * 100);

When you cast, cast type should be in brackets e.g. (cast type)value

C: filling an array with random numbers in a range

As mentioned in comments , if you want number in range 1 to 10 :

array0[i]= rand()%10 + 1;

I suggest int array1[10]={0}; instead of this loop:

for(int i = 0 ; i < 10 ; i++)
{
  array1[i]= 0;
}

and here is complete code with printing:

int main()
{
int n;
printf("How many elements in array?:");
scanf("%d",&n);
int array0[n];
for(int i = 0 ; i < n ; i++)
{
  array0[i]= rand()%10 + 1;//for your range
}

int array1[10]={0};

/*for(int i = 0 ; i < 10 ; i++)
{
  array1[i]= 0;
}*/
int index;
for(int i = 0 ; i < n ; i++)
{
  index = array0[i];
  array1[index-1]++;
}
    for (int i = 0; i < 10; i++)
    {
        printf("number %d appears:%d\n", i + 1, array1[i]);
    }
}

also as @Ardent Coder said add srand(time(NULL)); bfeore rand() to generate different random numbers at different runtimes.

How to fill an array with different random numbers from 1000 to 9999?

You should call Randomize() before the actual loop:

Randomize();
for i := 1 to n do
begin
  T[i] := Random(9000) + 1000;
end;

Randomize initializes the random number generator

Fill an array with random values and then count them

Instead of using array_fill. You generate a random values and put them in array.

<?php
 $arr = [];
 for($i=0;$i<10;$i++){
   $arr[$i] = rand(1, 5);
 }
 $a = array_count_values($arr);
 $b = array_keys($a, min($a));
 print_r($b);
?>

Create an array with random values

Here's a solution that shuffles a list of unique numbers (no repeats, ever).

for (var a=[],i=0;i<40;++i) a[i]=i;

// http://stackoverflow.com/questions/962802#962890
function shuffle(array) {
  var tmp, current, top = array.length;
  if(top) while(--top) {
    current = Math.floor(Math.random() * (top + 1));
    tmp = array[current];
    array[current] = array[top];
    array[top] = tmp;
  }
  return array;
}

a = shuffle(a);

If you want to allow repeated values (which is not what the OP wanted) then look elsewhere. :)

How do I fill an array with random numbers such that they are different?

I hope this will help, I tried to stay in the line with your first draft and what you were going for, just to note that this should work for an N length array. I changed the conditions on your second while to check the conditions before placing the value- and now you don't need to go over the set array and check and update the values.

you can also go another way as was commented here and just fill the array with values with help of one aux array to check each value is used only once and then randomly swap the indexes under the conditions.

I wrote this down but I didn't run tests- so make sure you understand whats going on and upgrade it to your needs. I do recommend using only one aux array, easy on memory and less whiles and checks.

GOOD LUCK

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

#define N 16
#define C 4

void FillArray(int *sites, int length) {
/*these aux arrays will keep track if an index was fill of if a value was used*/
int checkarrIndex[N] = { 0 };
int checkarrVal[N] = { 0 };

int i, cnt = 0, full=0; /*full is when all index are filled */
int *p = sites;
while (cnt < C) {
    i = rand() % length;
    if (checkarrIndex[i] == 0) /* checkarrIndex will let you know if an index has been gvin a value*/
    {
        ++checkarrIndex[i]; /*now  checkarrIndex[i] will be one so this index is now not valid for placement next time*/
        p[i] = -1;
        ++full;/*at the end of this while full will equal 4*/
        cnt++;
    }

}
while (full < length) /*here you need to draw a random index and a random value for it, 
                  not just a random value for a fixed index like you did, if I got this wrong just
                  go over the free indexes and place a rand value one at a time in the same manner*/
{
    int index; /*will store new random index */
    int value; /*will store new random value */
    index = rand() % N;
    value = rand() % N;/*max value is 15*/
    while(checkarrIndex[index]!= 0) /*check if this index was already placed */
    {
        index = rand() % N; /*try a another one */
    }
    /*I made this while loop to check all the con before filling the array */
    while(checkarrVal[value]!= 0 || p[value]== index || index == value) /*check if this value was already used  or if p[i]=j&&p[j]=i cond happens and make sure p[a] != a*/
    {
        value = rand() % N; /*try a another one */
    }
    ++checkarrIndex[index];/*set index as used */
    ++checkarrVal[value];/*set value as used */
    p[index] = value;
    ++full; /*another place was filled */


  }
}
static void PrintArray(int* arr, size_t size)
{
    int i = 0 ;
    for (i = 0 ; i< size; ++i)
    {
        printf("%d| ", arr[i]);
    }
    printf("\n");
}
int main(void)
{
    int array[N] = {0};
    FillArray(array, N);
    PrintArray(array, N);
    return 0;
}

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