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digitalmars.D.learn - inout based on this reference?

reply "Simen kjaeraas" <simen.kjaras gmail.com> writes:
Is it possible to make a function return dependent upon the constancy of  
the this reference?

i.e. this:

struct foo {
    int n;
    inout( ref int ) getn( ) inout {
        return n;
    }
}


-- 
Simen
Aug 29 2010
next sibling parent "Steven Schveighoffer" <schveiguy yahoo.com> writes:
On Sun, 29 Aug 2010 17:36:24 -0400, Simen kjaeraas  
<simen.kjaras gmail.com> wrote:


 Is it possible to make a function return dependent upon the constancy of  
 the this reference?

 i.e. this:

 struct foo {
     int n;
     inout( ref int ) getn( ) inout {
         return n;
     }
 }



Yes.  That is the bread-and-butter usage for inout.

Although inout is currently horribly broken.

-Steve
Aug 30 2010
prev sibling next sibling parent "Steven Schveighoffer" <schveiguy yahoo.com> writes:
On Mon, 30 Aug 2010 09:08:11 -0400, Steven Schveighoffer  
<schveiguy yahoo.com> wrote:


 On Sun, 29 Aug 2010 17:36:24 -0400, Simen kjaeraas  
 <simen.kjaras gmail.com> wrote:


 Is it possible to make a function return dependent upon the constancy  
 of the this reference?

 i.e. this:

 struct foo {
     int n;
     inout( ref int ) getn( ) inout {
         return n;
     }
 }



 Yes.  That is the bread-and-butter usage for inout.

 Although inout is currently horribly broken.



I should point out, the function sig should be:

ref inout(int) getn() inout

inout is a type modifier, ref is a storage class.

-Steve
Aug 30 2010
prev sibling next sibling parent "Simen kjaeraas" <simen.kjaras gmail.com> writes:
Steven Schveighoffer <schveiguy yahoo.com> wrote:
0123456789012345678901234567890123456789012345678901234567890123456789012

 On Sun, 29 Aug 2010 17:36:24 -0400, Simen kjaeraas  
 <simen.kjaras gmail.com> wrote:


 Is it possible to make a function return dependent upon the constancy  
 of the this reference?

 i.e. this:

 struct foo {
     int n;
     inout( ref int ) getn( ) inout {
         return n;
     }
 }



 Yes.  That is the bread-and-butter usage for inout.

 Although inout is currently horribly broken.



So horribly broken that example does not work at all, or is that just me?

This is my code, verbatim:

//////////////////////////
module foo;

struct bar {
     int n;
     ref inout( int ) getN( ) inout {
         return n;
     }
}

void main( ) {
}
//////////////////////////

And it gives this message:

foo.d(5): Error: inout on return means inout must be on a parameter as
well for inout inout(int)()

-- 
Simen
Aug 30 2010
prev sibling parent "Steven Schveighoffer" <schveiguy yahoo.com> writes:
On Mon, 30 Aug 2010 13:49:20 -0400, Simen kjaeraas  
<simen.kjaras gmail.com> wrote:


 Steven Schveighoffer <schveiguy yahoo.com> wrote:
 0123456789012345678901234567890123456789012345678901234567890123456789012

 On Sun, 29 Aug 2010 17:36:24 -0400, Simen kjaeraas  
 <simen.kjaras gmail.com> wrote:


 Is it possible to make a function return dependent upon the constancy  
 of the this reference?

 i.e. this:

 struct foo {
     int n;
     inout( ref int ) getn( ) inout {
         return n;
     }
 }



 Yes.  That is the bread-and-butter usage for inout.

 Although inout is currently horribly broken.



 So horribly broken that example does not work at all, or is that just me?

 This is my code, verbatim:

 //////////////////////////
 module foo;

 struct bar {
      int n;
      ref inout( int ) getN( ) inout {
          return n;
      }
 }

 void main( ) {
 }
 //////////////////////////

 And it gives this message:

 foo.d(5): Error: inout on return means inout must be on a parameter as
 well for inout inout(int)()



Yes, horribly broken means does not work in almost all cases :)  You  
should defer using inout until it is fixed.  As of now, it's unusable  
(Walter is aware and has it on his TODO list).

Please vote for bug http://d.puremagic.com/issues/show_bug.cgi?id=3748

-Steve
Aug 30 2010