Why is it that wcout << “”; is OK but wcout << string(); is not?

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我在风中等你
我在风中等你 2021-01-01 18:12
#include 
#include 

using namespace std;

int main()
{
    wcout << L\"Hello\";          // OK.
    wcout << wstring(L\"He         


        
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  • 2021-01-01 18:26

    For the first one, I'm guessing this overload is used:

    template< class CharT, class Traits >
    basic_ostream<CharT,Traits>& operator<<( basic_ostream<CharT,Traits>& os, 
                                             const char* s );
    

    Where wstream is essentially a basic_ostream<wchar_t>.

    For why string("Hello") doesn't work, it's simply because there is no conversion from string to wstring, nor an overload of operator<< provided.

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  • 2021-01-01 18:28

    This is dictated by § 27.7.3.6.4 of the C++11 Standard, where the following two overloaded operators (among others) are specified:

    template<class charT, class traits>
    basic_ostream<charT,traits>& operator<<(
        basic_ostream<charT,traits>& out, 
        const charT* s
        );
    
    template<class charT, class traits>
    basic_ostream<charT,traits>& operator<<(
        basic_ostream<charT,traits>& out, 
        const char* s
        );
    

    The last overload deals explicitly with char-based C-strings. This means that even for instantiations of the basic_ostream<> class template with the argument wchar_t there will be one overload which will deal with narrow char strings.

    Moreover, per § 27.7.3.6.4/5:

    Padding is determined as described in 22.4.2.2.2. The n characters starting at s are widened using out.widen (27.5.5.3). The widened characters and any required padding are inserted into out. Calls width(0).


    On the other hand, the statement wcout << string("Hello"); does not compile because string does not have an implicit conversion to const char*, and because there is no overload of operator << that would insert a string built with one character type into an output stream with a different underlying character type.

    In Standard terms (see § 21.4.8.9), here is how the definition of the overloaded operator << looks like for std::string:

    template<class charT, class traits, class Allocator>
    basic_ostream<charT, traits>& operator<<(
        basic_ostream<charT, traits>& os,
        const basic_string<charT,traits,Allocator>& str
        );
    

    As you can see, the same template parameter charT is used to instantiate both basic_ostream and basic_string.

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