• Paul (14/14) Apr 30 2015 When compiled on a 64 bit machine, this line
• bearophile (9/11) Apr 30 2015 ".length" returns a size_t, and 0 is an int. uniform() probably
• Paul (6/18) May 01 2015 Thank you for the explanation, it makes perfect sense despite

"Paul" <paul example.com> writes:

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

"bearophile" <bearophileHUGS lycos.com> writes:

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

"Paul" <paul example.com> writes:

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