I come from a pascal background and we could use:
div    integral division operator
/      floating point division operator

So my question is, how does D force floating point division on integrals?
At the moment i do this, but i was hoping for an easier way:

int n1 = 10, n2 = 2;
float f = cast(float)(cast(float)(n1 / n2));

int n1 = 10, n2 = 2;
float f = (n1+0.0f)/n2;

Casting n1 to float would also work, but I hope the compiler is smart enough to optimize away the plus expression.

Damian:

> I come from a pascal background and we could use:
> div    integral division operator
> /      floating point division operator

Two operators for the two different operations is a design better
than C, that is bug-prone.


> So my question is, how does D force floating point division on integrals?
> At the moment i do this, but i was hoping for an easier way:
>
> int n1 = 10, n2 = 2;
> float f = cast(float)(cast(float)(n1 / n2));

That's not good, it performs an integer division, followed by two
float casts.

Note: float is useful only if you have many of them, or if you
pass/return pairs of them. A single float is not so useful.

A solution:

int n1 = 10, n2 = 2;
const f = n1 / cast(double)n2;

Using type inference is useful, as it doesn't hide an integer
result if your code is wrong.

Bye,
bearophile

On Thursday, 11 October 2012 at 15:21:01 UTC, bearophile wrote:
> Damian:
>
>> I come from a pascal background and we could use:
>> div    integral division operator
>> /      floating point division operator
>
> Two operators for the two different operations is a design better
> than C, that is bug-prone.
>
>
>> So my question is, how does D force floating point division on integrals?
>> At the moment i do this, but i was hoping for an easier way:
>>
>> int n1 = 10, n2 = 2;
>> float f = cast(float)(cast(float)(n1 / n2));
>
> That's not good, it performs an integer division, followed by two
> float casts.
>
> Note: float is useful only if you have many of them, or if you
> pass/return pairs of them. A single float is not so useful.
>
> A solution:
>
> int n1 = 10, n2 = 2;
> const f = n1 / cast(double)n2;
>
> Using type inference is useful, as it doesn't hide an integer
> result if your code is wrong.
>
> Bye,
> bearophile

Ah i see, thankyou for the explanation.