Question

In my program i declared a function prototype like :

void callToPrint(LPTSTR , LPVOID , DWORD , string )

But i get the following error due to this statement : error C2061: syntax error : identifier 'string'

There are other errors also in the code which tell that the function does not take 4 arguments.

(error C2660: 'callToPrint' : function does not take 4 arguments)

Why do i get this error ? And how can i fix them ?

My second question is :

• I have declared a variable nameofPrinter of type LPTSTR but when i write the statement getline( cin , nameOfPrinter ) , the error displayed is no instance of overloaded function getline matches the argument list. Then how can i receive the nameOfPrinter from the user ?

Answer 1

Your file needs to contain the following line:

#include <string>

The header file string contains the definition for the string class. Because that class is within the std namespace the function prototype needs to be:

void callToPrint(LPTSTR , LPVOID , DWORD , std::string );

Since you’re using LPTSTR in the prototype you must be using Visual C++. If your project is set to use Unicode charset instead of multi-byte char set you need to adapt your types accordingly.

For Unicode charset:

std::wstring nameOfPrinter;
std::getline( std::wcin , nameOfPrinter );

Or, you could declare the type of your strings as:

std::basic_string<TCHAR> nameOfPrinter;

Unfortunately, such a templated class does not exist for switching between cin and wcin. So you’ll have to resort to the preprocessor.

#if defined(UNICODE) || defined(_UNICODE)
  #define _tcin wcin
#else
  #define _tcin cin
#endif

std::basic_string<TCHAR> nameOfPrinter;
std::getline( std::_tcin , nameOfPrinter );