Re: "out" and "in out"

[Martin Krischik <krischik@users.sourceforge.net>]

zork wrote:

> now I get as a result "CUT", and i dont understand why i get this result.
> Doesnt the "out" specify that its initial value isnt passed in via the
> parameter? But it seems to be passed in the above. In fact the "out" is
> acting like an "in out". I am a little confused. Could someone shed some
> light on this?

No it means: Does not *need* to be passed in. However, if using a pointer
instead of the value gives better performance the compiler will do that.
Also there are types which for technical reasons need to be passed using a
pointer.

What you have done comes under "undefined beahavior". If you used:

type s_type is new string (1 .. 3);

s : s_type = "CAT";

procedure modify ( s1 : out s_type) is
begin
ￜￜￜs1(2)ￜ:=ￜ'U';
end modify; 

things *might* be different on standart CPU's since s_type is smaller then
32 bit. BTW: String is indefinite and size therefore undefined. 

With Regards

Martin

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