Brace-init-list and assignments

#include <iostream>

struct int_wrapper
{
    int i;

    int_wrapper(int i = 0) : i(i) {}
};

struct A
{
    int_wrapper i;
    int_wrapper j;

    A(int_wrapper i = 0, int_wrapper j = 0) : i(i), j(j) {}
};

int main()
{
    A a;

    a = {3};

    // error: no match for ‘operator=’ (operand types are ‘A’ and ‘int’)
    // a = 3;

    std::cout << a.i.i << std::endl;

    return 0;
}

I know that isn't allowed more than one user-conversion when doing an implicit conversion, but, why with a brace-init-list the double user-conversion works?