Question

I have an (uncommented…) source file which I’m trying to understand.

static const Map *gCurMap;
static std::vector<Map> mapVec;

then

auto e = mapVec.end();
auto i = mapVec.begin();
while(i!=e) {
    // ...
    const Map *map = gCurMap = &(*(i++));
    // ...
}

I don’t understand what &(*(i++)) does. It does not compile when just using i++, but to me it looks the same, because I’m “incrementing” i, then I’m requesting the value at the given address and then I’m requesting the address of this value?!

Answer 1

Not at all. &*x is the same as operator&(operator*(x)), which can be anything you want.

It is only true for pointer types like T * p that &*p is the same as p. But C++ has user-defined types and overloadable operators.

The dereference operator (*) is typically overloaded for iterators to return a reference to the container element. The effect of the ampersand operator (&) on the container element is up to the class author; if you want to take the address unconditionally, you should use std::addressof(*i++) (from your fa­vou­rite header<memory>).