Re: Implicit conversion error
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
Implicit conversion error
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
Re: Implicit conversion error
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
