通过ostream输出C数组

Question

我正在尝试使用iostream输出C数组.

对于int的数组,我写了这样的代码

template <size_t N>
ostream& operator<< (ostream& os, const int (&x)[N])
{
    for(int i=0; i<N; i++)
        os<<x[i]<<",";
    return os;
}
int main()
{
    int arr[]={1,2,3};
    cout<<arr<<endl;
    return 0;
}

它工作得很好.

然后,我将它推广到更多类型(如字符,浮点数等),所以我更新原始版本如下

template <class T, size_t N>
ostream& operator<< (ostream& os, const T (&x)[N])
{
    for(int i=0; i<N; i++)
        os<<x[i]<<",";
    return os;
}

主要功能没有改变,但是,这次,当我编译它时,发生了错误.

In function `std::ostream& operator<<(std::ostream&, const T (&)[N]) [with T = int, long unsigned int N = 3ul]':
a.cpp:15:   instantiated from here
a.cpp:9: error: ambiguous overload for `operator<<' in `(+os)->std::basic_ostream<_CharT, _Traits>::operator<< [with _CharT = char, _Traits = std::char_traits<char>]((*((+(((long unsigned int)i) * 4ul)) + ((const int*)x)))) << ","'
/usr/lib/gcc/x86_64-redhat-linux/3.4.5/../../../../include/c++/3.4.5/bits/ostream.tcc:121: note: candidates are: std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT, _Traits>::operator<<(long int) [with _CharT = char, _Traits = std::char_traits<char>] <near match>
/usr/lib/gcc/x86_64-redhat-linux/3.4.5/../../../../include/c++/3.4.5/bits/ostream.tcc:155: note:                 std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT, _Traits>::operator<<(long unsigned int) [with _CharT = char, _Traits = std::char_traits<char>] <near match>
/usr/lib/gcc/x86_64-redhat-linux/3.4.5/../../../../include/c++/3.4.5/bits/ostream.tcc:98: note:                 std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT, _Traits>::operator<<(bool) [with _CharT = char, _Traits = std::char_traits<char>]

我怎样才能解决这个问题？谢谢你的任何建议.

Answer 1

已经存在一个operator << (const char*)与模板不一致的重载.

您可以使用SFINAE限制模板以排除char:

template <class T, size_t N,
     typename = typename std::enable_if<!std::is_same<char, T>::value>::type>
ostream& operator<< (ostream& os, const T (&x)[N])