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From: gwyn@smoke.BRL.MIL (Doug Gwyn)
Newsgroups: comp.std.c
Subject: Re: Is a CAST (double) -> (int) guaranteed to truncate ?
Message-ID: <13334@smoke.BRL.MIL>
Date: 10 Jul 90 02:47:42 GMT
References: <1431@tub.UUCP>
Organization: U.S. Army Ballistic Research Laboratory, APG, MD.
Lines: 24

In article <1431@tub.UUCP> heim@tub.UUCP (Heiner Marxen) writes:
>It occurs to me that most C programmers assume that the cast operator
>from double to int truncates the value (rounds towards 0), i.e. the above
>program is expected to print " -1 -1 1 1".  But, is this behaviour
>guaranteed/enforced by the definition of C?

The C standard requires that conversion of a floating type to an integer
type discard the fractional part, i.e. truncate toward zero.  Pre-ANSI C
implementations differ in their treatment of such cases.  I even found
one compiler recently that rounded on
	double d = 0.9; long i = d;
but truncated on
	double d = 0.9; long i = (long)d;

>A similar question applies to integer division.  Consider the C program
>	main() { printf("%d %d\n", 4/3, 5/3); }
>Is "1 1" its only legal output, or may it print "1 2"?

Integer division of positive operands has always been defined in C as
truncating toward zero (discarding remainder).  Behavior with negative
operands has deliberately been allowed to vary, in order to permit the
simple use of a hardware-divide instruction in the generated code.
The C standard DOES require that i/j and i%j obey a certain definite
algebraic relationship, though: (i/j)*j+i%j==i.