Does an odd number always return floor when divided with a remainder?

Question

Can I be sure that an odd number in C++ should always return floor of the result when divided in such a way that there is a remainder or are there any exceptions to this? I mean:

int x = 5;
x = x/2;
cout<<x;      //2

By 'floor' do you mean [std::floor](http://en.cppreference.com/w/cpp/numeric/math/floor)? — cgmb, Apr 18 '13 at 16:53

Koushik Shetty · Answer 1 · 2013-04-18T17:21:39.103

5

yes. you can be sure of that in c++

ISO/IEC N3485(working draft) says in 5.6.4

The binary / operator yields the quotient, and the binary % operator yields 
the remainder from the division of the ﬁrst expression by the second.
   If the second  operand of / or % is zero the behavior is undeﬁned. 
For integral operands the / operator yields the algebraic quotient with any 
fractional part discarded;81 if the quotient a/b is representable in the type 
of the result, (a/b)*b + a%b is equal to a; otherwise, the behavior of both 
a/b and a%b is undeﬁned.

edited Apr 18 '13 at 17:21

answered Apr 18 '13 at 16:39

Koushik Shetty

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well its free,this draft :-) – Koushik Shetty Apr 18 '13 at 16:50
That's because it's not a language standard. In this case, of course, the numbering and text are the same. – Lightness Races in Orbit Apr 18 '13 at 16:50
yes will buy one and post from that hence forth. is there a major difference by the way? – Koushik Shetty Apr 18 '13 at 16:51
Ah, N3485 is in fact the _current_ working draft, from October, i.e. pre-pre-pre-C++14. See how it's not immediately clear, unless you quote an actual standard? :) – Lightness Races in Orbit Apr 18 '13 at 16:54
oh thanks. quite a lot of differences there. have to be more careful henceforth. thanks for the suggestion. – Koushik Shetty Apr 18 '13 at 16:58
What does "algebraic quotient" mean? – brian beuning Apr 18 '13 at 17:40
"**algebraic quotient**" is nothing but the rules defined in algebra for division yielding quotient and remainder, which is nothing but `(numerator - remainder)/denominator`(all elements are whole numbers). this is our everyday division. – Koushik Shetty Apr 18 '13 at 18:25

score 2 · Answer 2 · edited May 23 '17 at 12:31

2

Integer division is handled as a floor operation in C/C++.

You get 2 in the above example since the real answer 2.5 can't be represented.

Some more verbose answers here

edited May 23 '17 at 12:31

Community

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answered Apr 18 '13 at 16:29

Guvante

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score 2 · Answer 3 · answered Apr 18 '13 at 16:48

Yes; division between integers is always integral division in C++:

[C++11 5.6/4]: The binary / operator yields the quotient, and the binary % operator yields the remainder from the division of the first expression by the second. If the second operand of / or % is zero the behavior is undefined. For integral operands the / operator yields the algebraic quotient with any fractional part discarded; if the quotient a/b is representable in the type of the result, (a/b)*b + a%b is equal to a.

Does an odd number always return floor when divided with a remainder?

3 Answers3