Out parameters and initialization

[Regan Heath]

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