Re: [C] Pointer to an array

Anand Hariharan said:

> On Wed, 14 May 2008 18:32:05 +0000, Richard Heathfield wrote:

<snip>

>> usr_data is converted to &usr_data[0]
> 
> Is that true?

Yes.

> I thought usr_data[0] implies a dereference,

You don't actually need to access the value of usr_data[0] (which is what 
"dereference" means) just to establish its address, which is what & does. 
(Indeed, without knowing its address, it's hard to see how you *could* 
evaluate it.)

> hence
> &usr_data[0] is better expressed as &(*(usr_data)) rather than usr_data.

I don't see why that's better. Surely a simple expression is better than a

complicated expression? We write n, not (n + 8) - 8. We write n, not (n * 
10) / 10. We write sqrt(n), not n/sqrt(n). For the more complicated 
expression to be worth writing, it has to achieve something useful. What 
does &(*(usr_data)) tell us that usr_data doesn't?

-- 
Richard Heathfield <http://www.cpax.org.uk>
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