For example, suppose you have these variables:
int i = 9;
int j = 7;
Depending on the implementation, the value of, (-i)/j, could be either –1 or –2. How is it possible to get these two different results?
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Keith Thompson
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Luis Averhoff
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@Raphael Yes if we were talking about C99, but a c book that I'm reading says that in C89 it's possible to get different results if you were to perform a calculation like this. – Luis Averhoff May 01 '15 at 17:36
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@LuisAverhoff , which book? – Arun A S May 01 '15 at 17:37
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@ArunA.S C_Programming_A_Modern_Approach_2nd_Ed – Luis Averhoff May 01 '15 at 17:37
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2To whoever voted to close this: How is it off-topic? I wouldn't be surprised if it's a duplicate, but it's certainly about programming. – Keith Thompson May 01 '15 at 17:43
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1@KeithThompson Well before I posted this question I didn't see any questions relating to this but it's probably buried deep inside the website. – Luis Averhoff May 01 '15 at 17:45
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@KeithThompson yes that's better. – Luis Averhoff May 01 '15 at 17:46
2 Answers
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Surprisingly the result is implementation defined in C89:
ANSI draft § 3.3.5
When integers are divided and the division is inexact, if both operands are positive the result of the / operator is the largest integer less than the algebraic quotient and the result of the % operator is positive. If either operand is negative, whether the result of the / operator is the largest integer less than the algebraic quotient or the smallest integer greater than the algebraic quotient is implementation-defined
However this was changed in C99
N1256 § 6.5.5/6
When integers are divided, the result of the / operator is the algebraic quotient with any fractional part discarded*
With a footnote:
* This is often called "truncation toward zero"
To clarify, "implementation defined" means the implementation must decide which one, it doesn't mean sometimes you'll get one thing and sometimes you'll get another (unless the implementation defined it to do something really strange like that, I guess).
Ryan Haining
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1"Implementation-defined" *also* means that the implementation must document its choice. – Keith Thompson May 01 '15 at 19:41
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I was reading the Ansi draft and I came across this sentence: "If the quotient a/b is representable, the expression (a/b)*b + a%b shall equal a". What does the word representable mean in this context? – Luis Averhoff May 01 '15 at 21:05
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@LuisAverhoff if the type has the capacity to accurately "be" something. 0.5 in not representable as an int, just as the result of 1/3 is not reperesentable as an int – Ryan Haining May 01 '15 at 21:07
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In C89, the result of division / can be truncated either way for negative operands. (In C99, the result will be truncated toward zero.)
The historical reason is explained in C99 Rationale:
Rationale for International Standard — Programming Languages — C §6.5.5 Multiplicative operators
In C89, division of integers involving negative operands could round upward or downward in an implementation-defined manner; the intent was to avoid incurring overhead in run-time code to check for special cases and enforce specific behavior. In Fortran, however, the result will always truncate toward zero, and the overhead seems to be acceptable to the numeric programming community. Therefore, C99 now requires similar behavior, which should facilitate porting of code from Fortran to C.
