Path: utzoo!utgpu!news-server.csri.toronto.edu!cs.utexas.edu!sun-barr!decwrl!borland.com!pete
From: pete@borland.com (Pete Becker)
Newsgroups: comp.lang.c++
Subject: Re: function parameters
Message-ID: <1991May28.184900.3692@borland.com>
Date: 28 May 91 18:49:00 GMT
References: <2569@m1.cs.man.ac.uk>
Organization: Borland International
Lines: 17

In article <2569@m1.cs.man.ac.uk> jk@cs.man.ac.uk (John Kewley ICL) writes:
>Why is this, apart from "because the syntax is defined like that", why is 
>a function with consts on its parameters not valid in place of one with
>variable as parameters?

	A function that takes const parameters promises not to modify those
parameters.  A function that takes non-const parameters makes no such promise.
It is legal to pass either a const or a non-const object to a function that
takes a const parameter, because the constness of the object is preserved.  It
is not legal to pass a const object to a function that takes a non-const 
parameter, because the function could modify the passed parameter.  In this
situation, the compiler will create a temporary and pass that as the
actual parameter, so the original object will not, in fact, be modified.  But
when you use a pointer to a function, there's no way for the compiler to know,
at the point of the function call, whether to generate that temporary.  So
the rule on function pointers has to be more strict: the parameter types
must agree completely.