How to return address of array from a function?

Question

can someone help me with this code.The code below has to accept input values for the array from the function 'ip' and print the array .

Address of array should be passed as input to the function .This is my code and i'm getting this error([Error] invalid conversion from 'int*' to 'int' [-fpermissive]).

#include<iostream>
using namespace std;
int ip(int *p);

int main()
{
int a[10];
int *ptr;
*ptr= ip(a);
for(int i=0;i<10;i++)
{
cout<<*(ptr+i)<<endl;
}

return 0;
}
int ip(int*p)
{
for(int i=0;i<10;i++)
{
cin>>*(p+i);
}
return a;
}

Answer 1

declare your function as a function pointer int *ip(int *p), that should work

Answer 2

You're God bruhh💯

Answer 3

Actually, there's no need to mess with return type since the nature of passing an array (unlike a simple variable) to another function involves decaying the array into a pointer with the proper type (the array lose its dimension and therefore the ability to get the size no longer possible). That means, `ip(a)` is the same as `ip(&a[0])` and what is being done inside the `ip` function would directly influence the actual array in the main(). So, as a conclusion, the above code can simply be revised as

#include<iostream>
using namespace std;

void ip(int *p);

int main()
{
int a[10];

ip(a);

for (int i = 0; i < 10; i++)
{
cout << *(a + i) << endl;
}

return 0;
}

void ip(int *p)
{
for (int i = 0; i < 10; i++)
{
cin >> *(p + i);
}
}