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Solved
Pointers in C++
Posted on 2011-03-01
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
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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6 Comments
Active 2 days ago
>> 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.
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.
Active 2 days ago
>> dereference whatever is found in the first slot of the array aVar.
Using the aforementioned equivalence, what you attempting to do is:
**(aVar + 1)
Using the aforementioned equivalence, what you attempting to do is:
**(aVar + 1)
//
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
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
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));
Active 2 days ago
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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