What is the difference between the following ways of printing a string?

std::cout << "hello";

// and

std::cout << std::string("hello");

They both appear to do the same thing.

CodePudding user response：

std::cout << "hello"; uses the operator<< overload dealing with null terminated character arrays. "hello", which is a const char[6], decay s into a const char* when passed as an argument to the operator<< overload:

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

std::cout << std::string("hello"); uses the operator<< overload for dealing with strings. std::string is a typedef for std::basic_string<char, std::char_traits<char>, std::allocator<char>>.

template <class CharT, class Traits, class Allocator>
std::basic_ostream<CharT, Traits>&
    operator<<(std::basic_ostream<CharT, Traits>& os,
               const std::basic_string<CharT, Traits, Allocator>& str);

CodePudding user response：

Case 1

Here we consider the statement:

std::cout << "hello";

In the above statement "hello" is a string literal of type const char[6] which decays to a const char* due to type decay. This const char* is then passed to the non-member overload of operator<< given below:

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

Case 2

Here we consider the statement:

std::cout << std::string("hello");

In the above statement the expression std::string("hello") creates a temporary std::string object. Now since the std::string has overloaded operator<<(given below), that overloaded version of operator<< will be called and its body will be executed.

template <class CharT, class Traits, class Allocator>
std::basic_ostream<CharT, Traits>&
    operator<<(std::basic_ostream<CharT, Traits>& os,
               const std::basic_string<CharT, Traits, Allocator>& str);