Modulus operator changes

Question

$5.6/4 in C++03 states- "If both operands are nonnegative then the remainder is nonnegative;if not, the sign of the remainder is implementation-defined⁷⁴⁾.

where Note 74 is

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

C++0x states -

$5.6/4- "For integral operands the / operator yields the algebraic quotient with any fractional part discarded;⁷⁹ if the quotient a/b is representable in the type of the result, (a/b)*b + a%b is equal to a.

Note 79 says

This is often called truncation towards zero.

So I have two questions:

• Can someone explain this concept of 'truncation towards zero'?

• Is modulus with negative operands a defined behavior in C++0x?

Answer 1

Truncation towards zero means converting a real number to an integer by selecting the next integer nearest zero. Equivalently, you write the number down, and ignore everything after the decimal point, whether the number is positive or negative.

Consider 11/4 = 2.75 -- if you truncate this towards zero, you get 2.

consider -11/4 or 11/-4 = -2.75 -- if you truncate this towards zero, you get -2.

It's important for some maths operations that (a/b)*b + a%b == a. If we must make this equation hold, and we also accept that integer division truncates towards zero, then we can deduce the operation of the % operator as follows:

a == 11, b == 4:
a/b == 2, 2*4 + a%b == 11, therefore a%b == 3.

a == -11, b == 4:
a/b == -2, -2 * 4 + a%b == -11, therefore a%b == -3.

a == 11, b == -4:
a/b == -2, -2 * -4 + a%b == 11, therefore a%b == 3.

a == -11, b == -4:
a/b == 2, 2 * -4 + a%b == -11, therefore a%b == -3.

It might be confusing at first, but C++0x is defining the behaviour of the a%b operator using the equation (a/b)*b + a%b == a. This equation holds even for negative numbers, so a%b is defined for negative numbers.

Answer 2

a) Consider (±5)/(±3) -> ±1. On the number line:

  (-5)/3                                     5/3
     5/(-3)                               (-5)/(-3)
      =                                       =
    -1.66 --> -1                       1 <-- 1.66
       v       v                       v       v
-  +  -  -  -  +  -  -  -  +  -  -  -  +  -  -  -  +  -
   |           |           |           |           |
  -2          -1           0           1           2

the rounding is thus towards zero.

b) Yes. Since a/b is now defined for all a and b (except b == 0), and (a/b)*b + a%b == a, there is only 1 unique solution for a%b, so the % operator is also well-defiend for all a and b (except b == 0).