Re: why use -> (not .) with pointers?

On Thu, 30 Jun 2005 07:14:37 -0700, Suman wrote:

>
>
> Richard Tobin wrote:
> [snip]
>>
>> I think the answer is just that they are different operations so C has
>> different names for them. a->b is equivalent to (*a).b.
> Don't you think that was what I meant, albeit in a roundabout way?
> You don't I'll assume for now :)
>
> <i_think_remotely_O(n)T(opic)>
> In any case, this fires my curiosity. What if we did have the _same_
> operator? Would the compiler not have to do extra work?

Not to any measurable degree. It has to validate the type of the left hand
operand either way.

Lawrence
.

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