Passing arguments into functions - in, out, inout, const, and contracts
Jason House
jason.james.house at gmail.com
Sat Feb 10 22:09:33 PST 2007
I believe that everything (except maybe fundamental types such as int)
are passed by reference. I'd expect "in" parameters to be considered
const, but I know that member functions can't be declared const.
It seems to me that in/out/inout has no effect on how the language
operates. I'm wondering how those affect the language. Does "in"
impose a contract that the ending value must equal the starting value?
I also haven't thought of any good way to differentiate out from inout
from the compiler's standpoint (obviously, it makes sense to users).
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