Bernard - 1 month ago 6x

C++ Question

Consider the following code (in C++11):

`int a = -11, b = 3;`

int c = a / b;

// now c == -3

C++11 specification says that division with a negative dividend is rounded toward zero.

It is quite useful for there to be a operator or function to do division with rounding toward negative infinity (e.g. for consistency with positive dividends when iterating a range), so is there a function or operator in the standard library that does what I want? Or perhaps a compiler-defined function/intrinsic that does it in modern compilers?

I could write my own, such as the following (works only for positive divisors):

`int div_neg(int dividend, int divisor){`

if(dividend >= 0) return dividend / divisor;

else return (dividend - divisor + 1) / divisor;

}

But it would not be as descriptive of my intent, and possibly not be as optimized a standard library function or compiler intrinsic (if it exists).

Answer

I'm not aware of any intrinsics for it. I would simply apply a correction to standard division retrospectively.

```
int div_floor(int a, int b)
{
int res = a / b;
int rem = a % b;
// Correct division result downwards if up-rounding happened,
// (for non-zero remainder of sign different than the divisor).
int corr = (rem != 0 && ((rem < 0) != (b < 0)));
return res - corr;
}
```

Note it also works for pre-C99 and pre-C++11, i.e. without standarization of rounding division towards zero.

Source (Stackoverflow)

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