When compiled on a 64 bit machine, this line

int r = uniform(0, mobs.length);

gives me an error:
Error: cannot implicitly convert expression (uniform(0, mobs.length)) of type ulong to int

but it compiles ok on a 32 bit machine.

I thought it was the expression on the righthand side returning a ulong which won't 'fit' in an int but if I substitute a numerical value instead of trying to get the length, eg uniform(0, 5) it compiles.

Why is that?

(mobs is an array of structs)

TIA

Paul

Paul:

> When compiled on a 64 bit machine, this line
>
> int r = uniform(0, mobs.length);

".length" returns a size_t, and 0 is an int. uniform() probably decides to unify those types to a size_t. A size_t is 32 bit on 32 bit machines and 64 bits on 64 bit machines. But D "int" is always a 32 bit signed integer. D allows implicit assignment of a 32 bit size_t to int but not a 64 bit size_t to an int. I agree that it's a bit of a mess.

Bye,
bearophile

On Thursday, 30 April 2015 at 22:24:15 UTC, bearophile wrote:
> Paul:
>
>> When compiled on a 64 bit machine, this line
>>
>> int r = uniform(0, mobs.length);
>
> ".length" returns a size_t, and 0 is an int. uniform() probably decides to unify those types to a size_t. A size_t is 32 bit on 32 bit machines and 64 bits on 64 bit machines. But D "int" is always a 32 bit signed integer. D allows implicit assignment of a 32 bit size_t to int but not a 64 bit size_t to an int. I agree that it's a bit of a mess.
>
> Bye,
> bearophile

Thank you for the explanation, it makes perfect sense despite being a bit of a surprise. (I should have worked this out for myself but I haven't figured out how to use the documentation properly yet!).

Paul