Is there any special C function to get XOR of all element of array?-CodePudding

I have an array like this [1, 0, 1, 1, 0, 0, ...].

How can i get result of this expression: 1 XOR 0 XOR 1 XOR 1 XOR... without loop?

CodePudding user response：

In C 14:

std::accumulate(arr.begin(),
                arr.end(),
                0,
                std::bit_xor<void>())

CodePudding user response：

You could create a function template that takes an array and forwards to a helper function using make_index_sequence - which then uses a fold expression to "unpack" the array.

This does not require any loop since the expression will be created at compile time.

#include <utility>

namespace detail {
template<class T, std::size_t... I>
auto Xor_helper(const T& arr, std::index_sequence<I...>) {
    // make a fold expression of the whole array:
    return (... ^ arr[I]);      // arr[0] ^ arr[1] ^ ...
}
} // namespace detail

template<class T, size_t N>
auto Xor(const T(&arr)[N]) {
     return detail::Xor_helper(arr, std::make_index_sequence<N>{});
}

And you could call it with your array like so:

int main() {
    int arr[] = {0xf, 0x1, 0x3};

    std::cout << std::hex << Xor(arr) << '\n';
}

Output:

You can see what becomes of these templates @ cppinsights. Just press the ▶ button and you'll see the above Xor(arr) become return (arr[0UL] ^ arr[1UL]) ^ arr[2UL];

CodePudding user response：

It worked: std::accumulate(arr.begin(), arr.end(), arr[0], std::bit_xor());