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Vieux 15/01/2008, 19h40   #5
Luis Angel Fdez. Fdez.
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Par défaut Re: Why pA->foo() works in this code?

El Tue, 15 Jan 2008 08:15:40 -0800, tragomaskhalos escribió:

Hi, again.

Thank you all for your anwsers.

>> void MakeA(A *pA) { pA = new A; }

>
> This just sets pA and then forgets it value: you mean void MakeA(A*& pa)
> { pA = new A; }


That's what we thought, but when we saw it working (or doing something
it seemed right) we were surprised.


>> Â And... why ((A*)NULL)->foo() works too?
>>

>
> This came up recently, see:
> http://groups.google.co.uk/group/com...thread/thread/
> a807522e71a60c8e/f4afa72f427ed638?hl=en


Thank you. I was searching something about that, but I couldn't find
anything.

Bye.

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