stdio line-streaming revisited

[Derek Parnell]

On Thu, 29 Mar 2007 22:40:34 -0700, Walter Bright wrote:

> kris wrote:
>>    Cout.opCall("Hello, ").opCall(Cin.get);
> 
> Given:
> 	a.f(b,c);
> can be equivalently written as:
> 	f(a,b,c);
> 
> we can transform:
> 	Cout.opCall("Hello, ").opCall(Cin.get);
> into:
> 	opCall(opCall(Cout,"Hello, "), Cin.get);
> 
> And it is transformed that way, because the optimizer/back end knows 
> nothing about member functions. They're just more function calls.
>
> The nested opCall will get called before the outer opCall, but otherwise 
> the order is undefined (and will change depending on the function 
> calling convention, whether it is a virtual call or not, etc.). 

Now that makes sense. I can see now why the W&A show is concerned. In fact,
the call-chaining example seems to flesh out to ...

   opCall(opCall(Cout,"Hello, "), get(Cin));

which means that either the 'get' could be called first or the inner
'opCall' could be called first. One can never be sure.

This would mean that the sequence example given by Kris ...

    Cout ("Please enter your name: ").flush;
    Cout ("Hello, ") (Cin.get);

gets transformed in to 

    flush( opCall( Cout, "Please enter your name: "));
    opCall( opCall( Cout, "Hello, "), get(Cin));

The first statement will end up displaying the text on the console, leaving
the cursor to the right of the ':' character. Now if the 'get' is called
next, the user types in her name, say "mavis" and the screen will show ...

  Please enter your name: mavis
  _

and then the inner 'opCall' is run which will show (assuming a flush) ...

  Please enter your name: mavis
  Hello, mavis
  _


However, if the inner 'opCall' is called first the screen will show ...

  Please enter your name: Hello, _

and the user types in her name and we get ...

  Please enter your name: Hello, mavis
  mavis
  _


So the order is reasonable important to sort out.

Being the conservative procedural type of coder, I would have probably have
coded ...

    char[] userName;
    Cout ("Please enter your name: ").flush;
    userName = Cin.get();
    Cout ("Hello, ");
    Cout (userName);
    Cout.flush;

Actually, now that I think about it, I'd really do ...

    char[] userName;
    char[] myResponse;

    Cout ("Please enter your name: ");
    Cout.flush;
    userName = Cin.get();
    myResponse = std.string.format("Hello, %s\n", userName);
    Cout.( myResponse );
    Cout.flush;

That is of course, if I'd really used stdin/stdout in that manner at all
<G>

-- 
Derek
(skype: derek.j.parnell)
Melbourne, Australia
"Justice for David Hicks!"
30/03/2007 3:57:25 PM