Re: Pass by reference

[dan.r.mcleran@seagate.com (Dan McLeran)]

> Good Ada compiler will do what is best. So "array (1 .. 2) of Character" is
> likely to be passed by value while "array (1 .. 10_000) of Integer" is
> probably passed by reference. And some Ada compilers might even use pass by
> register.

Not sure what you mean by 'pass by register'. Maybe each value of a
2-element array are loaded into registers and then read by the calling
function?

> 
> This might sound strange to a C, C++ or C# programmer. Like giving up
> control. There are things humans can do better then computers but there are
> also things computers can do better then humans - and counting is is among
> those. The compiler knows much better how many registers are available, how
> large the type is, how long it takes to copy and so on.

It's not too foreign of idea to me because I use a C++ library to do
essentially the same thing. The Boost Call Traits Library is a
template library that chooses the optimal parameter passing convention
based on type. Since the C++ project I have been working on makes
heavy use of templates, I am used to leaving it up to
someone/something else to choose the best calling convention.

> 
> You say what you need (in, out, in out) and the compiler will choose the
> optimal way to fullfill you which. 

This is what I want to solidify in my brain. Alot of people posting
replies seem to say that the compiler has more freedom than the RM
specifies. If most Ada compilers are conforming, I would assume that
the compiler only has freedom for unspecified types. By-reference and
by-values types must be passed as stated in RM 6.2.

> 
> > A 2nd part to my question is: Does Ada automatically pass tagged types
> > by reference no matter what mode the parameter is specfied (in, in
> > out, or out)? The language RM 6.2 seems to suggest this is so. If so,
> > does this mean that there is no way to pass a tagged type by value? I
> > believe C# works this way without programmer intervention.
> 
> They are in deed allways passed by reference. That's done to enable
> dispaching. Rermember: all native procedures of a tagged type are "virtual".
> 
> Two ways around that behavior:
> 
> 1st: for any tagged type X there is indefinite type X'Class. X'Class is an
> indefinite but untagged type so RM 6.2 does not apply here. 
> 
> 2nt: RM 6.2 does not apply to a non native procedure. That is a procedure 
> in another package.
> 
> With Regards
> 
> Martin