C ++整数楼层函数

Question

I want to implement greatest integer function. [The "greatest integer function" is a quite standard name for what is also known as the floor function.]

int x = 5/3;

My question is with greater numbers could there be a loss of precision as 5/3 would produce a double?

EDIT: Greatest integer function is integer less than or equal to X. Example:

4.5 = 4
4 = 4
3.2 = 3
3 = 3

What I want to know is 5/3 going to produce a double? Because if so I will have loss of precision when converting to int.

Hope this makes sense.

Answer 1

You will lose the fractional portion of the quotient. So yes, with greater numbers you will have more relative precision, such as compared with 5000/3000 .

However, 5 / 3 will return an integer, not a double. To force it to divide as double, typecast the dividend as static_cast<double>(5) / 3 .

Answer 2

Integer division gives integer results, so 5 / 3 is 1 and 5 % 3 is 2 (the remainder operator). However, this doesn't necessarily hold with negative numbers. In the original C++ standard, -5 / 3 could be either -1 (rounding towards zero) or -2 (the floor), but -1 was recommended. In the latest C++0B draft (which is almost certainly very close to the final standard), it is -1, so finding the floor with negative numbers is more involved.

Answer 3

5/3将始终产生1（整数），如果你做5.0 / 3或5 / 3.0，结果将是双倍。

Answer 4

Since in C and C++, as others have said, / is integer division, it will return an int. in particular, it will return the floor of the double answer... (C and C++ always truncate) So, basically 5/3 is exactly what you want.

It may get a little weird in negatives as -5/3 => -2 which may or may not be what you want...

Answer 5

As far as I know, there is no predefined function for this purpose. It might be necessary to use such a function, if for some reason floating-point calculations are out of question (eg int64_t has a higher precision than double can represent without error)

We could define this function as follows:

#include <cmath>

inline long
floordiv (long num, long den)
{
  if (0 < (num^den))
    return num/den;
  else
    {
      ldiv_t res = ldiv(num,den);
      return (res.rem)? res.quot-1 
                      : res.quot;
    }
}

The idea is to use the normal integer divison, but adjust for negative results to match the behaviour of the double floor(double) function. The point is to truncate always towards the next lower integer, irrespective of the position of the zero point. This can be very important if the intention is to create even sized intervals.

Timing measurements show that this function here only creates a small overhead compared with the built-in / operator, but of course the floating point based floor function is significantly faster....