Deny std::vector from deleting its data

Question

I have the following case:

T* get_somthing(){
    std::vector vec; //T is trivally-copyable
    //fill vec
    T* temp = new T[vec.size()];
    memc         


        

Answer 1

Ok, don't try this at home, it's not nice. It's more an experiment than an actual answer.

(Well, the requirements/design is not nice either: "Play stupid games, win stupid prizes")

in your cpp:

#define private public               // good luck for the code review
#define protected public
#include                     // (must be the first occurence in the TU)
#undef private                       // do not abuse good things...
#undef protected

template
T* my_release(std::vector& v){
    std::vector x;                // x: the local vector with which we mess around
    std::swap(x, v);                 // the given vector is in an OK, empty state now.
    T* out = x._M_impl._M_start;     // first, get the pointer you want

    // x will be destructed at the next '}'. 
    // The dtr only use _M_start and _M_finish, make sure it won't do anything.
    x._M_impl._M_start = nullptr;    
    x._M_impl._M_finish = nullptr;

    // no need to say, the internal state of 'x' is bad, like really bad...
    // also we loose the capacity information, the actual allocator... 
    // -> good luck with memory leaks...

    return out;
}

// usage example
int main(){
    std::vector vi{1,2,3,4,5,6,7,8,9};
    auto n = vi.size();
    int* pi = release(vi);
    for(size_t i=0; i

prints 1, 2, 3, 4, 5, 6, 7, 8, 9,