Question

c++ Write two versions of a function that returns a dynamic array consisting of all prime numbers less than or equal to the int parameter n. You may declare the dynamic array to have length n, even though this is a bit inefficient. In one version of the function, return the number of primes via a separate reference variable. In the other, use 0 as a sentinel value to signify the end of the prime list. See below for the function prototypes.

int* smallPrimes(int n, int& size); // number of primes returned via size ref variable
int* smallPrimes(int n); // sentinel value indicates end of prime list

Implement both of the functions in main with two input values from the user.

Answer 1

#include <iostream>

using namespace std;

int* smallPrimes(int n, int& size) // number of primes returned via size ref variable

{

int *arr= new int[n];

int index=0,count=0;

//iterating in range

for(int i=2;i<=n;i++){

count=0;

//checking if is prime

for(int j=2;j<i;j++){

if(i%j==0)

count++;

}

if(count==0)

arr[index++]=i;

}

//adding the size

size=index;

return arr;

}

int* smallPrimes(int n){ // sentinel value indicates end of prime list

int *arr= new int[n];

int index=0,count=0;

//iterating in range

for(int i=2;i<=n;i++){

count=0;

//checking if is prime

for(int j=2;j<i;j++){

if(i%j==0)

count++;

}

if(count==0)

arr[index++]=i;

}

//adding sentinel value as 0

arr[index]=0;

return arr;

}

int main(){

int size;

int *p=smallPrimes(20,size);

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

cout<<p[i]<<" ";

cout<<endl;

p=smallPrimes(30);

for(int i=0;p[i]!=0;i++)

cout<<p[i]<<" ";

}

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