No. Extension methods require an instance variable (value) for an object. You can however, write a static wrapper around the ConfigurationManager interface. If you implement the wrapper, you don't need an extension method since you can just add the method directly.
public static class ConfigurationManagerWrapper
{
public static ConfigurationSection GetSection( string name )
{
return ConfigurationManager.GetSection( name );
}
.....
public static ConfigurationSection GetWidgetSection()
{
return GetSection( "widgets" );
}
}
If you just want to pass a std::string to a function that needs const char* you can use
std::string str;
const char * c = str.c_str();
If you want to get a writable copy, like char *, you can do that with this:
std::string str;
char * writable = new char[str.size() + 1];
std::copy(str.begin(), str.end(), writable);
writable[str.size()] = '\0'; // don't forget the terminating 0
// don't forget to free the string after finished using it
delete[] writable;
Edit: Notice that the above is not exception safe. If anything between the new call and the delete call throws, you will leak memory, as nothing will call delete for you automatically. There are two immediate ways to solve this.
boost::scoped_array
boost::scoped_array will delete the memory for you upon going out of scope:
std::string str;
boost::scoped_array<char> writable(new char[str.size() + 1]);
std::copy(str.begin(), str.end(), writable.get());
writable[str.size()] = '\0'; // don't forget the terminating 0
// get the char* using writable.get()
// memory is automatically freed if the smart pointer goes
// out of scope
std::vector
This is the standard way (does not require any external library). You use std::vector, which completely manages the memory for you.
std::string str;
std::vector<char> writable(str.begin(), str.end());
writable.push_back('\0');
// get the char* using &writable[0] or &*writable.begin()
Best Answer
A
constobject is alwaysstatic.