CODEWITHSUNDEEP
c++c++11
John
John

Reputation: 2326

Integer Division in C++11

I was noticing some wording changes to section 5.6 for C++11. (I'm looking at the draft C++ standard N3242, dated 2011-02-28.) The new (draft) standard includes the sentence:

"For integral operands the / operator yields the algebraic quotient with any fractional part discarded;"

This statement is not in 5.6 of the 03 standard (ISO-IEC-14882-2003), but I don't think this is a change, is it? This is how C and C++ has worked for years unless I've lost my mind (which may have happened anyway).

Upvotes: 7

Views: 2523

Answers (2)

PlasmaHH
PlasmaHH

Reputation: 16046

Almost. In C++03 the sign of the remainder for % (in which terms both were specified) was unspecified, as such rounding could go away from zero in certain situations too. Compare with the C++03 footnote:

According to work underway toward the revision of ISO C, the preferred algorithm for integer division follows the rules defined in the ISO Fortran standard, ISO/IEC 1539:1991, in which the quotient is always rounded toward zero.

In practice however, this almost never made any difference.

Upvotes: 2

Lightness Races in Orbit
Lightness Races in Orbit

Reputation: 385144

You're not going mad.

A footnote to 5.6/4 said:

[C++03 footnote 74]: According to work underway toward the revision of ISO C, the preferred algorithm for integer division follows the rules defined in the ISO Fortran standard, ISO/IEC 1539:1991, in which the quotient is always rounded toward zero.

In C++11 this behaviour is explicitly required rather than being "preferred"; the change is listed in the compatibility section:

[C++11: C.2.2]:
Change: Specify rounding for results of integer / and %
Rationale: Increase portability, C99 compatibility.
Effect on original feature: Valid C++ 2003 code that uses integer division rounds the result toward 0 or toward negative infinity, whereas this International Standard always rounds the result toward 0.

Upvotes: 10

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