mod.std.c Digest V5#7

[Orlando Sotomayor-Diaz]

From: Orlando Sotomayor-Diaz (The Moderator) <cbosgd!std-c>


mod.std.c Digest            Sun, 21 Apr 85       Volume 5 : Issue   7 

Today's Topics:
                              #else JUNK
                      modulus, remainder, and %
----------------------------------------------------------------------

Date: Mon, 15 Apr 85 15:31:33 pst
From: nsc!turtlevax!ken at ihnp4.uucp (Ken Turkowski)
Subject: #else JUNK
To: ihnp4!cbosgd!mark at nsc.uucp

>A colleague of mine recently noted their sadness with the (November, 1984)
>draft standard's outlawing of source such as
>	#ifdef VERBOSE
>		printf("Hello, world.\n");
>	#endif VERBOSE
>which, according to the draft, must have its last line changed to (for example)
>	#endif /* VERBOSE */
>to be legal.  Such code is now successfully dealt with on many systems.

The problem with this is that it removes the ability to structure pre-processor
directives.  For example,

#ifdef VERBOSE
# ifdef DEBUG
# else !DEBUG
# endif DEBUG
#else !VERBOSE
#endif VERBOSE

Indicates clearly which #ifdef is ended with each #endif, regardless of the
fact that the stuff after the #else and #endif is ignored.

------------------------------

Date: Mon, 15 Apr 85 15:31:33 pst
From: nsc!turtlevax!ken at ihnp4.uucp (Ken Turkowski)
Subject: modulus, remainder, and %
To: ihnp4!cbosgd!mark at nsc.uucp

>The % operator is called the "mod" operator in C, but it does not compute
>modulus for signed numbers in any C implementation I know of.  When I really
>want mod, I have to write (assuming b is unsigned):
>
>		a%b>0?a%b:a%b+b
>
>Would it break anything to really define % as mod?

What's the problem?  Both of them are equal (mod b).   :->

Seriously, the % operator is defined as the remainder of a division such that:

	(a / b) * b + (a % b) == a

Most implementations define the remainder to have the same sign as the
quotient, which has the property that:

	(-a) / b == -(a / b)

Forcing the remainder to be positive would mean that quotient may be one less
than it is in most implementations.  I think that the objection would not be
to defining % to yield a positive number, but to the change in the property
of quotients.

------------------------------

End of mod.std.c Digest - Sun, 21 Apr 85 11:19:48 EST
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