Out parameters and initialization

[Unknown W. Brackets]

If that is the meaning, some symbol should be used to indicate that 
there is a reference being made.

I don't think, personally, it makes any sense to say that in is a 
replacement for what might be passed.  It is much more like an 
initializer, which would never behave in that way.  To me, it is unclear 
and unlike the rest of the language to do as you suggest.

Your suggestion separates the logic of parameter defaulting from 
initialization completely, but they share the same syntax (as they share 
again with assignment.)  I think this is, at best, unclear and clumsy. 
At worst, it is obviously inconsistent and confusing.

But that is more of a philosophical debate, and that will be for Walter 
to answer, ultimately.  My point is that there is a bug here, nonetheless.

Regardless of what you are saying, why should my initial example 
compile?  It should either work as initialization, or not compile.  It 
should not compile and do nothing.  This is also unclear and inconsistent.

-[Unknown]


> On Sat, 25 Feb 2006 16:21:23 -0800, Unknown W. Brackets 
> <unknown at simplemachines.org> wrote:
>> That does not make it logical or consistent.  It makes it confusing 
>> and inconsistent.  Utility is not at all my concern.
> 
> Ok, so lets look at the consistency of this.
> 
>> In other words, you're essentially saying that this makes sense:
>>
>> var2 = 2;
>> var1 = var2;
>> var1 = 1;
>> assert(var2 == 1);
> 
> No, he's saying that this makes sense:
> 
> void foo(inout int var1) { var1 = 1; }
> 
> int var2 = 2;
> foo(var2);
> assert(var2 == 1);
> 
> Remember, 'inout' aliases the variable you pass, the parameter 'var1' is 
> another name for the variable 'var2'.
> 
> 
> So, when you write:
> 
> void foo(inout int var1 = 7)
> 
> what are you saying?
> 
> Lets start with what a default parameters means, in this case:
> 
> void foo(int var1 = 7) {}
> 
> says, if I do not get passed a variable, pass 7, in other words:
> 
> foo();
> 
> is called as:
> 
> foo(7);
> 
> Agreed?
> 
> 
> Therefore:
> 
> void foo(inout int var1 = 7)
> 
> says, if I do not get passed a variable, pass 7, meaning:
> 
> foo();
> 
> is called as:
> 
> foo(7);
> 
> Agreed?
> 
> 
> Now, as well all know you cannot pass '7' as inout because you cannot 
> alias '7'.
> 
> What I am trying to show here is that default parameters are behaving 
> both logically and consistently, they always mean: "If I do not get 
> passed a variable, pass X" where X is the default parameter value supplied.
> 
> Further, 'inout' is behaving both logically and consistently, in all 
> cases it aliases the parameter which is passed.
> 
> 
> The behaviour of 'in'+'default parameter' is not the same as 
> 'inout'+'default parameter' but why should it be? IMO one is an apple 
> and the other an orange.
> 
> Regan