c++ - 通过ostream输出一个C数组

Question

我正在尝试使用 iostream 输出 C 数组。

对于整数数组，我这样写代码

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;
}

而且效果很好。

然后，我将其泛化为更多类型(如字符、 float 等)，因此我将原始版本更新如下

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;
}

main函数没有变化，但是这次编译的时候出现了错误。

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>]

我该如何解决这个问题？感谢您提出任何建议。

Answer 1

operator << (const char*) 已经存在重载这与您的模板不明确。

您可以使用 SFINAE 限制您的模板以排除 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])