Is there an easy, efficient and correct (i.e. not involving conversions to/from double) way to do floored integer division (like e.g. Python offers) in C#.
In other words, an efficient version of the following, that does not suffer from long/double conversion losses.
(long)(Math.Floor((double) a / b))
or does one have to implement it oneself, like e.g.
static long FlooredIntDiv(long a, long b)
{
if (a < 0)
{
if (b > 0)
return (a - b + 1) / b;
// if (a == long.MinValue && b == -1) // see *) below
// throw new OverflowException();
}
else if (a > 0)
{
if (b < 0)
return (a - b - 1) / b;
}
return a / b;
}
*) Although the C# 4 spec of the Division operator leaves it open whether OverflowException
is raised inside unchecked
, in reality it does throw (on my system) and the Visual Studio .NET 2003 version even mandated it throw:
If the left operand is the smallest representable int or long value and the right operand is –1, [..] System.OverflowException is always thrown in this situation, regardless of whether the operation occurs in a checked or an unchecked context.
Edit
The crossed out statements about checked
and unchecked
are all nice and well, but checked
is in fact only a compile time concept, so whether my function should wrap around or throw is up to me anyway, regardless of whether code calling the function is inside checked
or not.
In Python, we can perform floor division (also sometimes known as integer division) using the // operator. This operator will divide the first argument by the second and round the result down to the nearest whole number, making it equivalent to the math. floor() function.
With the correct cast the value is the same. the answer is yes.
If you imagine a room where 3 is on the ceiling and 2 is on the floor. 2.5 would fit in the middle. Floor division means the “//“ will always take the floor or the lower number.
Floor division is an operation in Python that divides two numbers and rounds the result down to the nearest integer. The floor division happens via the double-backslash (//) operator. r = a // b. r = a // b.
You can try this:
if (((a < 0) ^ (b < 0)) && (a % b != 0))
{
return (a/b - 1);
}
else
{
return (a/b);
}
Edit (after some discussions in comments below):
Without using if-else, I would go like this:
return (a/b - Convert.ToInt32(((a < 0) ^ (b < 0)) && (a % b != 0)));
Note: Convert.ToIn32(bool value)
also needs a jump, see implemention of the method:
return value? Boolean.True: Boolean.False;
Theoretically, it is not possible to calculate the division for a = long.MinValue
and b = -1L
, since the expected result is a/b = abs(long.MinValue) = long.MaxValue + 1 > long.MaxValue
. (Range of long is –9,223,372,036,854,775,808
to 9,223,372,036,854,775,807
.)
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