enable_if compilation question void = nullptr

Does anyone know why assigning type* = 0 doesn’t work, while type* = nullptr does? In both cases typedef void type. Thanks

#include <type_traits>
#include <iostream>

template <class T,
      typename std::enable_if<std::is_integral<T>::value>::type* = 0>
void do_stuff(T& t) {
    std::cout << "do_stuff integraln";
}

#if 0 // works
template <class T,
      typename std::enable_if<std::is_integral<T>::value>::type* = nullptr>
void do_stuff(T& t) {
    std::cout << "do_stuff integraln";
}
#endif


struct S {};
int main(int argc, char *argv[])
{
    int i = 1;
    do_stuff(i);
    return 0;
}

Compilation:

clang++ -pedantic -Wall -std=c++11 test190.cc && ./a.out
test190.cc:23:5: error: no matching function for call to 'do_stuff'
    do_stuff(i);
    ^~~~~~~~
test190.cc:6:6: note: candidate template ignored: substitution failure
      [with T = int]: null non-type template argument must be cast to template
      parameter type 'typename
      std::enable_if<std::is_integral<int>::value>::type *' (aka 'void *')
void do_stuff(T& t) {
     ^
1 error generated.