Algorithm to add sum of every possible xor-sum sub-array
After the first couple of lines the array $A$ contains the XOR of every prefix of the input numbers in binary. So a $1$ bit at position $k$ in $A[j]$ tells us that there are an odd number of input numbers in the range $0...j$ with $1$ at their position $k= \log p$.

For every $i,j$ s.t. $i \leq j$, if the $k$th bit of $A[i-1]$ is $1$ and the same bit in $A[j]$ is $0$ then there are an odd number of $1$s at bit-position $k$ of the input numbers in the range $i...j$. The same conclusion would be true if the $k$th bit was $0$ for $A[i-1]$ and $1$ for $A[j]$. Please observe that these are the only two situations where the number of ones at position $k$ is odd in the range $i...j$. 

Therefore the number of pairs $(A^k[i-1],A^k[j])=(0,1)$ or $(1,0)$ is equal to the number of ranges in the input such that their XOR at position $k$ is $1$. Every such range contributes $p=2^k$ to the total sum. In that code, $c$ is the number of ones and $(N-c)$ is the number of zeros at position $k = \log p$. The "$+1$" is to count for the ranges of length one and a $1$ in their $k$th position.