获取给定地址的 std::vector 元素的索引

Question

Let's say I have a std::vector and I get by some means the address of the n-th element. Is there a simple way (faster than iterating through the vector) to get the index at which the element appears, given the base address of my std::vector? Let's assume I'm sure the element is in the vector.

Answer 1

Since you know the element is within the vector, and vector guarantees that its storage is contiguous, you could do:

index = element_pointer - vector.data();

or

index = element_pointer - &vector[0];

Note that technically the contiguous guarantee was introduced in C++03, but I haven't heard of a C++98 implementation that doesn't happen to follow it.

Answer 2

distance( xxx.begin(), theIterator);

The above will only work for a vector::iterator. If you only have a raw pointer to an element, you must use it this way:

distance(&v[0], theElementPtr);

Answer 3

Yes - because a vector guarantees all elements are in a contiguous block of memory you can use pointer arithmetic to find it like so

#include <iostream>
#include <vector>

int main(int argc, char *argv[])
{
   std::vector<int> vec;

   for(int i=0; i<10; ++i)
   {
      vec.push_back(i);
   }

   int *ptr=&vec[5];
   int *front=&vec[0];

   std::cout << "Your index=" << ptr-front << std::endl;
   return 0;
}

Answer 4

On the way of learning, I have taken the following notes:

#include <iostream>
#include <vector>

int main() {
    std::vector<std::string> words={"This","is","just","a","trial!"};
    size_t i; //using the contiguous property of a vector:
    for (auto const& elem : words) {
        i = &elem - &*words.begin();// or
        i = &elem - &words.front();// or 
        i = &elem - words.data();// or
        i = std::addressof(elem) - std::addressof(words[0]); 
        if(std::addressof(elem) == &words.front())
          std::cout << elem <<" (at"<<&elem<<") relative to ("<< &words[0] << ") takes position @#"<<i<< std::endl;
        else std::cout << elem <<" (at"<<&elem<< ") takes position @#"<<i<< std::endl;
    }
    return 0;
}

Test run here . It is open to further study (or learn from masters) which one is the most secured/safe and/or most efficient approach.