Set with Custom String Class Problem

I have written a custom string class. I want to use STL set with it. I have overloaded operator < But still its giving me problem

error C2678: binary '=' : no operator found which takes a left-hand operand of type 'const     String' (or there is no acceptable conversion)
1>    could be 'String &String::operator =(const String &)'
1>           'String &String::operator =(const char *)'
1>          'String &String::operator =(const wchar_t *)'
1>          while trying to match the argument list '(const String, const String)'

I guess, It is asking for overloaded operator= (const String , const String)

But its im开发者_JS百科possible to create such an overloaded function

My String class is this

String ();
String (const char * pStr);
String (const long int pData);
String (const double  pData);
String (const int pData);
String (const wchar_t * pStr);
//Copy Constructors
String (const String& rhs);
String (const String& rhs, const int pStartIndex, const int pNumChar);

//Overloaded Operators
String & operator= (const String & rhs);
String & operator= (const char * rhs);

String & operator= (const wchar_t * rhs);
String   operator+ (const String& rhs);
//String &  operator+= (const char ch);
String & operator+= (const String& rhs);
friend bool operator== (const String& lhs, const String& rhs);

friend bool operator< (const String& lhs, const String& rhs) {

    return strcmp(lhs.vStr, rhs.vStr);
}

friend ostream& operator<< (ostream& ostr, String& rhs);

char & operator[] (int pIndex);
char   operator[] (int pIndex) const;

const char * String::Buffer () const;
wchar_t * GetTChar();

int String::GetLength () const;

~String ();

"no operator found which takes a left-hand operand of type 'const String'"

it seem you have an expression like

a=b;

where both a and b are const String.

You cannot assign to a const (although the compiler looks desperately seeking for an implementation of such an assignment)

OK, well I can only answer the question you've posed with the information you've given, and the answer is that this works for me.