How come when using the string object for variables and parameters you need to have #include <string>, but you don't need it for string literals? For example, you can say cout << "This is a string literal"; without #include <string>.
I am learning C++ using a Deitel Brothers book and came up with this question when learning character arrays.
A string literal is not a std::string object, it's an array of const char.
"This is a string literal" has the type const char[25].
In most situations – including this one – an array implicitly decays into a pointer to its first element, and there is an operator<< overload for const char*.
It's pretty confusing that "string" means several different things in C++, but after a while (and pulling of hair and gnashing of teeth) the intended meaning will be clear from context.
"foo"s to get string, string literal
#include <string>A string literal like "This is a string literal" is of type const char[25], not of type std::string. Statement cout << "This is a string literal" actually calls operator <<(ostream&, const char*), and the string literal parameter decays to type const char*. There isn't any std::string involved in this case.
operator <<forstd::ostream&andconst char *(which, btw, can sometimes rear unexpectedly).strings. They are constant arrays of characters.std::coutis designed to print arrays of characters as strings.