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From: bobduff@world.std.com (Robert A Duff)
Subject: Re: Uninitialized variables, Java example
Date: 1996/07/25
Message-ID: <Dv47x9.150@world.std.com>#1/1
X-Deja-AN: 170833382
references: <31EEACDA.64880EEB@sage.inel.gov> <Dv0qo4.JoE@world.std.com>
 <Dv1sME.1u4.0.-s@inmet.camb.inmet.com> <4t7g2q$ee0@goanna.cs.rmit.edu.au>
organization: The World Public Access UNIX, Brookline, MA
newsgroups: comp.lang.ada
Date: 1996-07-25T00:00:00+00:00
List-Id: <comp.lang.ada>


In article <4t7g2q$ee0@goanna.cs.rmit.edu.au>,
Richard A. O'Keefe <ok@goanna.cs.rmit.edu.au> wrote:
>Wrong.  If the problem is that the programmer knows that the default:
>case can never be executed and the compiler doesn't, then the programmer
>should TELL the compiler.  For example, in the same situation in C, one
>does
>
>	int x;
>
>	switch (e) {
>	    case ..: ... x = ..; break;
>
>	    case ..: ... x = ..; break;
>	    default: abort();
>	}
>
>I actually use a "shouldnt" macro (name stolen with thanks from Interlisp)
>which is like assert(0) but cannot be suppressed by NDEBUG.
>
>Now the compiler knows that path can't be traversed too.

No, it doesn't.  It's quite possible to write the above, and have it
blow up at compile time.

>Better still, a human reader knows that the original programmmer *thought*
>about the problem and didn't just accidentally leave out the default:
>And there is no unncessary initialisation anywhere.

Agreed.  I would say, "Now the compiler knows that THE PROGRAMMER THINKS
that path can't be traversed too."  And another programmer knows that
the first programmer thinks...

- Bob