Conversion from std::vector::iterator to void *

Question

I am trying to convert std::vector<T>::iterator to void *, but getting compiler error as wrong conversion. is there any way?

No, there's no way to do that. If you describe the problem you're trying to solve by doing this, you might get a working solution to the actual problem. (This looks like an [XY problem](http://meta.stackexchange.com/questions/66377/what-is-the-xy-problem)). — molbdnilo, May 02 '16 at 14:04
If the compiler is complaining then you are probably doing something very bad. — Martin York, May 02 '16 at 15:14

Peter K · Accepted Answer · 2016-05-03T13:17:14.260

4

Just get the pointer to the vector element with dereference:

vector<Type>::iterator i = ...;
void* data = &*i;

Or to vector data:

void* data = vec.data();

edited May 03 '16 at 13:17

answered May 02 '16 at 14:08

Peter K

3

Or just `void *data = static_cast(vec.data());` – ForceBru May 02 '16 at 14:09
1

@ForceBru or even drop the cast : every pointer is implicitly convertible to (but not from) `void*`. – Quentin May 02 '16 at 14:23
`struct evil { char* operator&() const { return nullptr; } }; std::vector`: `&` is unreliable in generic code. – Yakk - Adam Nevraumont May 02 '16 at 20:00
Yakk, see this question for discussion on generic addressof: http://stackoverflow.com/questions/6494591/how-can-i-reliably-get-an-objects-address-when-operator-is-overloaded – Peter K May 03 '16 at 13:16

1 Answers1