spectral

&a->b  is equal to &(a->n) ou (&a)->b

Discussion created by spectral on Mar 3, 2011
Latest reply on Mar 3, 2011 by eugenek

Hi,

 

I suspect that sometimes the following expression "&a->b" is evaluated as :

(&a)->b

 

... but it should be &(a->n) ? Right ?

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