I have a function that does rounding operation as shown below. It takes 64bit integer as input and gives 32bit integer as output. While converting, a factor of 0x40000000 is being added to the input. What is the reason behind it?

```
int rounder(long long int in)
{
INT64 out;
if ((in >> 32) == 0x7FFFFFFF)
out = in;
else
out = (INT64)0x40000000 + in;
out = out >> 31;
return (INT32)out;
}
```

Let's start with some smaller numbers, because they're easier!

Using conventional rounding, x.49999... or less should round down to x, x.50000... or more should round up to (x+1).

(There are lots of different rounding methods, but this is the one most learn at school.)

Whenever you do integer division (or conversion of a floating point value to an integer), you simply throw away the fractional part. Hence:

```
6/2 == 3.0 --> 3
5/2 == 2.5 --> 2
```

A neat 'trick' is to add half-the-divisor (1, in this case) before division. As if by magic, you get the right rounding! eg:

```
6/2 becomes (6+1)/2 == 7/2 == 3.5 --> 3
5/2 becomes (5+1)/2 == 6/2 == 3.0 --> 3
```

You can see why this works by looking at it this way:

```
5/2 becomes (5+1)/2 == 5/2 + 1/2
13/6 becomes (13+3)/6 == 13/6 + 3/6 == 13/6 + 1/2
```

You're adding half to the real answer. Anything less than x.5 will still be less than x+1 so will still round down, anything of x.5 or more will become x+1 or more so will round up.

Now to your actual question: This idea works with all divisors; you're shifting down by 31, which is the same as dividing by 2^31. So 'half-the-divisor' is 2^30, or 0x40000000.

Beware: as others have noted, this 'trick' only works for positive numbers (you need to subtract if it's negative, but it's a can of worms).

There is a lot to consider in this topic; it's not simple to get your head around. As ever, try some easy examples for yourself and see what happens.

The input appears to be a 64-bit fixed point number with 31 fraction bits. The 0x40000000 value is added to round the number up if it has a fractional part >= 0.5. The `if`

is used to avoid possible overflow when factoring in the rounding.

answered on Stack Overflow Feb 20, 2020 by 1201ProgramAlarm

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