Meaning of in, out and inout

[Sergey]

https://forum.dlang.org/post/17nwtnp4are5q$.1ddtvmj4e23iy.dlg@40tude.net

On Tuesday, 10 May 2005 at 01:06:14 UTC, Derek Parnell wrote:
> On Tue, 10 May 2005 00:30:57 +0000 (UTC), Oliver wrote:
>
>> Hello D-ers
>> 
>> The documentation is very short on the keywords in, out and 
>> inout.
>> Is is inout sth like a reference ? But then, what is in and 
>> what is out?
>
>
> in:
> The argument is preserved, such that when control returns to 
> the caller,
> the argument as passed by the caller is unchanged. This means 
> that the
> called function can do anything it likes to the argument but 
> those changes
> are never returned back to the caller. There is a bit of 
> confusion here
> when it comes to passing class objects and dynamic arrays. In 
> both these
> cases, a reference to the data is passed. Which means that for 
> 'in'
> references, the called function is free to modify the reference 
> data (which
> is what is actually passed) in the full knowledge that any 
> changes will
> *not* be returned to the caller. However, if you make any 
> changes to the
> data being referenced, that modified data is 'returned'. Which 
> means that,
> for example, if you pass a char[] variable, the reference will 
> be preserved
> but the data in the string can be changed.
>
> out:
> The argument is always initialized automatically by the called 
> function
> before its code is executed. Any changes to the argument by the 
> called
> function are returned to the caller. The called function never 
> gets to see
> the value of the argument as it was before the called function 
> gets
> control. The argument must a RAM item and not a literal or 
> temporary value.
>
> inout:
> The argument is passed to the called function without before 
> its code is
> executed. Any changes to the argument by the called function 
> are returned
> to the caller. In other words, the called function can see what 
> value was
> passed to it before changing it.  The argument must a RAM item 
> and not a
> literal or temporary value.
>
>
> Examples:
>
>   char[] a;
>   int    b;
>
>   void func_one(in char[] X, in int Y)
>   {
>       X[0] = 'a';   // Modifies the string contents.
>       X = "zxcvb";  // Modifies the string reference but is not 
> returned.
>
>       Y = 3; // Modifies the data but is not returned.
>   }
>
>
>   a = "qwerty";
>   b = 1;
>   func_one(a,b);
>
>   writefln("%s %d", a,b); // --> awerty 1
>
>   void func_two(out char[] X, out int Y)
>   {
>       X[0] = 'a';   // Modifies the string contents.
>       X = "zxcvb";  // Modifies the string reference.
>       if (b == 1)
>         Y = 3; // never executed because Y is always zero on 
> entry.
>       else
>         Y = 4; // Modifies the data.
>   }
>
>
>   a = "qwerty";
>   b = 1;
>   func_two(a,b);
>
>   writefln("%s %d", a,b); // --> zxcvb 4
>
>   void func_three(inout char[] X, inout int Y)
>   {
>       X[0] = 'a';   // Modifies the string contents.
>       X = "zxcvb";  // Modifies the string reference.
>       if (b == 1)
>         Y = 3; // Modifies the data.
>       else
>         Y = 4; // Modifies the data.
>   }
>
>
>   a = "qwerty";
>   b = 1;
>   func_two(a,b);
>
>   writefln("%s %d", a,b); // --> zxcvb 3

Thanks a lot for your explanation.
I started to learn D language recently and I have trouble with 
understanding some part of language.
I do your example on Linux and it works very well, especially 
func_rthee().
when I try to repeat code on Windows 10, I get errors:
    Error: cannot modify `inout` expression `X[0]`
    Error: cannot modify `inout` expression `X`
    Error: cannot modify `inout` expression `Y`
    Error: cannot modify `inout` expression `Y`
Very strange situation for me.
I expect next behavior of the parameters X and Y could get value 
and could change it on Windows but now I am not sure what is goin 
on.

PS. I am using DMD64 D Compiler v2.098.1-dirty on Debian 10 Linux 
and Windows 10.