Question

Write a C++ program which inputs an array of integers from user then pass array pointer to function which will return sum of whole array. Also do same action by using address-based argument of array to another function.

Note: In this you need to declare two factions: 1. With argument of pointer type. 2.With argument type of address type

Answer 1

Program:

#include <iostream>
using namespace std;
int pointertype(int *,int);
int addresstype(int a[],int);
int main()
{
int n,i,result,array[100];
cout<<"enter a value:";
cin>>n;
cout<<"enter "<<n<<" numbers"<<endl;
for(i=0;i<n;i++)
cin>>array[i];
result=pointertype(array,n);
cout<<"Sumo of the array using pointertype:"<<result<<endl;
result=addresstype(array,n);
cout<<"Sumo of the array using addresstype:"<<result<<endl;
return 0;
}
int pointertype(int *p,int n)
{
int i,sum=0;
for(i=0;i<n;i++)
sum=sum+p[i];
return sum;
}
int addresstype(int a[],int n)
{
int i,sum=0;
for(i=0;i<n;i++)
sum=sum+a[i];
return sum;
}

screenshot of the program
main.cpp #include iostream» 2 using namespace std; 3 int pointertype(int *,int); 4 int addresstype (int al],int); 5 int main() int n,i,result,array [100]; cout<<"enter a value:; 7 8 cout<<"enter "<<n<<" numbers" <<endl; 10 11 for(i-0;i<n;i++) 12 13 cin> array[i]; result-pointertype(array,n); cout<<"Sumo of the array using pointertype:"<<result<<endl; result-addresstype(array,n); cout<< "Sumo of the array using addresstype: "<<result<<endl; 15 16 17 18 19 return 0 input Sumo of the array using addresstype:15

20 int pointertype(int "p,int n) int i, sum-0; for(i-0;i<n;i++) sum-sum p[i]; return sum; 23 24 25 26 27 int addresstype(int a[],int n) 28- 29 30 for(i-0;i<n;i++) 31 32 int i, sum-0 sum-sum+a[i]; return sum

output:

enter a value: 5 enter 5 numbers 1 2 3 4 5 Sumo of the array using pointertype:15 Sumo of the array using addresstvpe: ..Program finished with exit code 0 Press ENTER to exit console