typo correction :
(*aptr).xptr = 5;
On Tue, Aug 2, 2011 at 10:24 PM, Aniruddha Bhattacharyya
<aniruddha.aot@xxxxxxxxx> wrote:
> 1) a .* xptr = 3 ;
> Above line's syntax is wrong.
> Should be
> *a.xptr = 3;
>
> 2) Also declare the structure member variables as
> int *xptr, *yptr ; (otherwise yptr will be of type int, not of int*)
>
>
> 3) aptr ->* xptr = 5 ;
> is also of wrong syntax.
> Should be
> *aptr->xptr = 5 ;
>
> Which is equivalent to
> *(aptr->xptr) = 5;
>
>
>
> the above code DOES NOT mean this one :
>
> (*aptr)->xptr // this one dereference aptr first, then
> does nothing to xptr
>
>
>
> Thanks.
>
> On Tue, Aug 2, 2011 at 10:01 PM, Shriramana Sharma <samjnaa@xxxxxxxxx> wrote:
>>
>> Hello -- long time since I posted on this list.
>>
>> I'm trying to understand the use of ->* but in compiling the attached program which I thought was straightforward I am getting the following errors:
>>
>> foo.cpp: In constructor ‘mystruct::mystruct()’:
>> foo.cpp:9:12: error: invalid conversion from ‘int*’ to ‘int’
>> foo.cpp: In function ‘int main()’:
>> foo.cpp:21:7: error: ‘xptr’ was not declared in this scope
>> foo.cpp:22:7: error: ‘yptr’ was not declared in this scope
>>
>> Please help.
>>
>> --
>> Shriramana Sharma
>
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