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Pointers in C++

Posted on 2011-03-01
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Last Modified: 2012-05-11
Hello experts.

void aaa(bbb* aVar)
{
      bbb myVar;
      
      myVar = aVar[1]; //compiles
      //myVar = *aVar[1]; //does not compile

}

This is my thinking process:

a. bbb * aVar

aVar is a pointer to address of bbb.

We can treat aVar as an array where
aVar[0] points to first address of bbb
aVar[1] points to second address of bbb, and so on.

So...

bbb = aVar[1] should make bbb equal to address of second
element of aVar

bbb = *aVar[1] should make bbb equal to value of second
element of aVar


Where do I get it wrong?

Thank you

panJames


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Question by:panJames
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käµfm³d   👽 earned 500 total points
ID: 35006367
>>  aVar is a pointer to address of bbb.

aVar is a pointer to the location of an object of type bbb. The indexing operation ( [n] ) is equivalent to saying

    *(somePtr + n)

What you are trying to do with this line:

    myVar = *aVar[1];

is dereference whatever is found in the first slot of the array aVar.
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by:käµfm³d 👽
ID: 35006395
>>  dereference whatever is found in the first slot of the array aVar.

Using the aforementioned equivalence, what you attempting to do is:

    **(aVar + 1)
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Author Comment

by:panJames
ID: 35006411
//
What you are trying to do with this line:

    myVar = *aVar[1];

is dereference whatever is found in the first slot of the array aVar.
//


why is it illegal?

panJames
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Expert Comment

by:Mistralol
ID: 35006934

Thats kinda of true but what you are really doing is this

**(avar + sizeof(bbb));


void aaa(bbb* aVar)
{
      bbb myVar;
     
      myVar = aVar[1]; //compiles
      //myVar = *aVar[1]; //does not compile

}

The first works because myVar is taking a copy of the data in aVar[1] from the array slot 1.
So its the same as memcpy(&myVar, &aVar[1], sizeof(myVar));







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by:käµfm³d 👽
ID: 35007123
My C-Fu is not that strong, but IIRC, unless aVar is a pointer to an array of pointers, array of arrays, or a pointer to a pointer, you will get an error.
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by:käµfm³d 👽
ID: 35007128
*with the above syntax  = )
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