带数据交换的std :: vector指针

Question

In the section of code below, what would be the resultant memory structure after the swap? Would there be a leak because they have swapped memory addresses underneath? Would it be fine because they did a deep copy? what if this code was stuck inside of a class and I was swapping a working buffer with a piece of dynamic memory?

#include <iostream>
#include <vector>

int main()
{
    std::vector<std::string> * ptr_str_vec =
        new std::vector<std::string>();
    ptr_str_vec->push_back("Hello");

    std::vector<std::string> str_vec;
    str_vec.push_back("World");

    ptr_str_vec->swap(str_vec);

    delete ptr_str_vec;
    //What would be the resulting structures?

    return 0;
}

EDIT: Posted slightly faulty code. Fixed the errors.

Answer 1

When a vector is created, the underlying continuous data block used by vector is by default created from heap. In your case, since you didn't supply allocator, default one is used.

int main()
{
    std::vector<std::string> * ptr_str_vec =
        new std::vector<std::string>(); // #^&! *ptr_str_vec is allocated from heap. vector's data block is allocated from heap.
    ptr_str_vec->push_back("Hello");    // #^&! "hello" is copied onto heap block #1

    std::vector<std::string> str_vec;   // #^&! str_vec is allocated from stack. vector's data block is allocated from heap.
    str_vec.push_back("World");         // #^&! "world" is copied onto heap block #2

    ptr_str_vec->swap(str_vec);         // #^&! swap is fast O(1), as it is done by swapping block #1 and #2's address. No data copy is done during swap.

    delete ptr_str_vec;                 // #^&! delete ptr_str_vec as well as heap block #2.
    //What would be the resulting structures? /

    return 0;                           // #^&! delete str_vec as well as heap block #1
}

Answer 2

The values in each vector will be swapped http://www.cplusplus.com/reference/vector/vector/swap/

I see no memory leak (other than the one you get when your program ends at the end of main, since you aren't deleting your pointer), your ptr_str_vec pointer does not change, only the data inside the vector that it points to changes

Answer 3

Assuming that you are already familiar with swap , is there any reason that you haven't set it up so you can test the output to see what it does yourself? This will be the quickest way to assure yourself that you know exactly what it is doing and if your use of it is appropriate.

In this case, the resulting structures are simply that ptr_str_vec points to a vector containing a std::string("World") and str_vec is a vector containing a std::string("Hello") . Your example suffers from many faults in answering your question, in particular because you only have a single element in each vector (and thus the vectors are equal length), and because the elements are the exact same size (and thus the vectors occupy approximately equivalent memory segments). In a running instance of your full project, it is very likely that none of these conditions are true.