Allocating 8 bytes of memory dynamically

Question

I want to create an integer array that can hold 8 bytes of data, I have used this code

int n =50; 
long long *buffer = new long  long [n];

the size of long long is 8 bytes but

sizeof(buffer)

only gives 4 bytes is it that the heap doesn't allow more than 4 bytes allocation? there is something that I have missed here but don't know what it is.

That's the size of the pointer, not the size of the allocated memory - there is no way of finding that out. — , May 11 '18 at 16:19
`sizeof` gives the size of the pointer, not of the memory allocated. In C++ most of the array problems go away if you use a `std::vector` instead. — Bo Persson, May 11 '18 at 16:20
ah I see, that's basic pointer stuff that I should have known thank you guys for clearing it up. — Skeptical Ant, May 11 '18 at 16:23
cost of allocated memory 8B is relatively high (far, far more than 8B) , suggest design in different way — Jacek Cz, May 11 '18 at 16:25
@BoPersson I must use an array specifically because of supervisor other wise i would have used a vector or a list even. — Skeptical Ant, May 11 '18 at 16:30

score 3 · Accepted Answer · answered May 11 '18 at 16:21

3

sizeof (x) gives the number of bytes consumed by x. In your case x is the pointer to the allocation. If the allocation succeeds, you know that it allocated the requested size (or possibly a little more).

answered May 11 '18 at 16:21

wallyk

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Allocating 8 bytes of memory dynamically

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