Who can understand doctors?

We all use jargon. If we used it to help us communicate better then more people would understand us not less. As it goes our language is becoming more and more specialised with fewer and fewer people understanding us.

I heard two MA students in English from different Cambridge University colleges talking about English usage . Grabbing a phrase one of them had uttered I asked what they had said and what it meant.

"'Meta-ontological values' but it doesn't really mean anything specific." Said one.

"Sort of beyond a metaphor I think, but it is true we don't actually know what each other is on about.. We are both from different colleges" Said the other. After that we all got drunk and then started making sense.

The cause of jargon seems to be either:

• a desire to exclude others
• a desire to be thought of as more intelligent than we actually are
• a need to make our knowledge seem more obscure or precious
• a desire to form a clique
• simple laziness
• a desire to produce an brief, exact linguistic construct enabling clarification of status issues.

Isn't that interesting?

I would welcome it if you send me bits of interesting jargon you come across.

With C++ 2.1, when an operator++ or operator- is declared as a member function with no parameters, or as a nonmember function with one parameter, it only overloads the prefix operator ++ or operator -. You can only overload a postfix operator++ or operator- by defining it as a member function taking an int parameter or as a nonmember function taking one class and one int parameter. For example add the following lines to the previous code:

Turbo C++ for Windows Programmer's Guide

The reason for the size argument is that classes derived from X inherit the X::operator new. The size of a derived class object may well differ from that of the base class.

You can overload a prefix or postfix unary operator by declaring a nonstatic member function taking no arguments, or by declaring a non-member function taking one argument. If @ represents a unary operator, ~a x and x@ can both be interpreted as either x.operator@() or operator@(x), depending on the declarations made. If both forms have been declared, standard argument matching is applied to resolve any ambiguity.

Under C++ 2.0, an overloaded operator++ or - is used for both prefix and postfix uses of the operator.

I teach people this stuff and I don't understand it either.

Love and peace