Error in sum of array using recursion (Compilation and runtime) - c++

Question

I was asked to do sum of numbers in an array using recursion where sample input would be provided and the compiler gave me the following errors (note, I tried three ways and got the corresponding errors). Can someone please explain to me what I was doing wrong here?
Error for Code 1
In file included from Runner.cpp:2:0:
Solution.h: In function 'int sum(int*, int)':
Solution.h:7:32: error: invalid conversion from 'int' to 'int*' [-fpermissive]
totalsum = totalsum + sum(n+1,n-1);
^
Solution.h:1:5: note: initializing argument 1 of 'int sum(int*, int)'
int sum(int input[], int n) {
^
Error for code 2
Solution.h: In function 'int sum(int*, int)':
Solution.h:5:24: error: invalid conversion from 'int' to 'int*' [-fpermissive]
return input[0] + sum(n+1,n-1);
^
Solution.h:1:5: note: initializing argument 1 of 'int sum(int*, int)'
int sum(int input[], int n) {
^
Error for code 3
Runtime error
Code 1
int sum(int input[], int n) {
if (n<=0)
{
return 0;
};
int totalsum;
totalsum = totalsum + sum(n+1,n-1);
return totalsum;
}
Code 2
if (n<=0)
return 0;
return input[0] + sum(n+1,n-1);
Code 3
if (n <= 0)
return 0;
return (sum(input, n +1) + input[n - 1]);

Answer 1

In this function implementation
int sum(int input[], int n) {
if (n<=0)
{
return 0;
};
int totalsum;
totalsum = totalsum + `sum(n+1,n-1)`;
return totalsum;
}
the first parameter declared like int input[] has the type int *. However instead of the array you are passing the integer n + 1 as an argument in the recursive call
sum(n+1,n-1)
The same problem exists in this code snippet
if (n<=0)
return 0;
return input[0] + sum(n+1,n-1);
In the second code snippet it seems there is access memory beyond the array
return (sum(input, n +1) + input[n - 1]);
^^^^
And moreover the call has an infinite recursion because the second argument is always increased: n + 1.
The function can be written like
long long int sum( const int a[], size_t n )
{
return n == 0 ? 0 : a[0] + sum( a + 1, n - 1 );
}
Another approach is the following
long long int sum( const int a[], size_t n )
{
return n == 0 ? 0 :
( n == 1 : a[0] ? sum( a, n / 2 ) + sum( a + n / 2, n - n / 2 ) );
}
Here is a demonstrative program
#include <iostream>
long long int sum( const int a[], size_t n )
{
return n == 0 ? 0
: ( n == 1 ? a[0] : sum( a, n / 2 ) + sum( a + n / 2, n - n / 2 ) );
}
int main()
{
int a[] = { 1, 2,3, 4,5, 6, 7, 8, 9, 10 };
const size_t N = sizeof( a ) / sizeof( *a );
std::cout << sum( a, N ) << '\n';
return 0;
}
Its output is
55

Answer 2

For error for code 1, you are using an integer variable where it is expecting an array. To correct this, you would need to make n an index of some array. so the call would be something like sum(index[n+1], n-1) . Reason being is that arrays are pointers, pointers are arrays. so unless you have a pointer to an int variable, it will not work.