Question

How to inhabit the int array num with random numbers using a function that you then invoke from the main method.

I already created a function that generates an array of 5 random integers.

/* int i;

  int arrayName[5];

    for(i= 0;i < 6; i++ ){

           //type arrayName[arraysize];

           arrayName[i] = rand()% 200;

      printf("array[%d] = %d\n",i,arrayName[i]);

*/

// use rand() % some_number to generate a random number
int num[5];
int main(int argc, char **argv)
{
// hint
// function invocation goes here

return 0;
}

Answer 1

Basically here you want to store the randomly generated numbers from the function into the int num array. So in order to do that, there should be a function return type and you should return the int array. In C, functions cant return arrays. But there are 3 ways in which we can return array. I am using static array to do that. You can also use dynamic allocated array or struct to do that. Let us see that in the following code.

#include <stdio.h>
#include<stdlib.h>
int* generate() //return type is pointer
{
     int i;
     static int arrayName[5]; //declare it as static
    for(i= 0;i < 6; i++ ){
           arrayName[i] = rand()% 200; //generate random integers and store it in the arrayName
          //printf("array[%d] = %d\n",i,arrayName[i]);
    }
    return arrayName; ///return arrayName
}
int* num;
int main(int argc, char **argv)
{
    int i;
    num=generate();//invoke the function here. Now the values of arrayName will be stored in int array pointer
    
    //optional. i used for loop here to check the array values
    for(i= 0;i < 6; i++ ){
      printf("array[%d] = %d\n",i,num[i]);//now print the values 
    }
    //optional
   

    return 0;
}

I am also attaching the output and code screenshot for your reference.

Output and code screenshot:

#Please don't forget to upvote if you find the solution helpful. Feel free to ask doubts if any, in the comments section. Thank you.