declaration of void variable shadows template parameter in fold expression

Question

Is the following g++'s behaviour a bug?

#include <utility>
#include <type_traits>

template< std::size_t ...i >
constexpr
bool f(int k, std::index_sequence< i... >)
{
    int j = (std::size_t((void(i), 1)) + ...) * k;
    return 0 < j;
}

static_assert(f(3, std::make_index_sequence< 3 >{}));

Gives an error message:

main.cpp: In function constexpr bool f(int, std::index_sequence<i ...>):

main.cpp:11:33: error: declaration of void i shadows template parameter

int j = (std::size_t((void(i), 1)) + ...) * k;
                            ^

main.cpp:7:11: note: template parameter i declared here

template< std::size_t ...i >
          ^~~

Live example.

Changing from void to static_cast< void > makes error to cease.

Multiplication by k from the left side also makes error to cease.

std::size_t here is to avoid this g++'s bug.

I know, that declaration int (i), is equivalent to int i, in a couple of contexts, but not in lhs part of built-in comma operator. Moreover, it is a void variable declaration.