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From: Pete.Becker@sunbrk.FidoNet.Org (Pete Becker)
Newsgroups: comp.lang.c++
Subject: Re: function parameters
Message-ID: <675518012.71@sunbrk.FidoNet>
Date: 28 May 91 23:49:00 GMT
Sender: Usenet@sunbrk.FidoNet.Org
Lines: 20


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.

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