How to prevent unique pointers from overlapping

Question

I'm trying to build a timer in C++. In my Timer class there are two Date objects that hold a std::unique_ptr<struct tm> pointer. After I std::move the second unique_ptr in the second Date object in the following piece of code, it points to the same memory address of the first one, effectively making the two objects represent the same time, even though they should be different because of the duration offset.

    using namespace std;
    time_t localTime = time(nullptr);
    unique_ptr<struct tm> currentTime = static_cast<unique_ptr<struct tm>>(localtime(&localTime));
    startDate = Date(std::move(currentTime));

    time_t endTime = time(nullptr) + duration;
    unique_ptr<struct tm> timerEndTime = static_cast<unique_ptr<struct tm>>(localtime(&endTime));
    endDate = Date(std::move(timerEndTime));

The Date constructor being called is this:

Date::Date(std::unique_ptr<struct tm> time) : date(std::move(time)) {}

What am I doing wrong?

Answer 1

It is unsurprising that the two objects overlap, because that is how std::localtime is supposed to work:

std::tm* localtime( const std::time_t *time );

Return value

pointer to a static internal std::tm object on success, or null pointer otherwise.

- See std::localtime on cppreference

If you create a std::unique_ptr that wraps the results of std::localtime, you will attempt to free an object with delete which was never allocated with new, and that is undefined behavior. Write:

std::time_t localTime = std::time(nullptr);
std::tm* currentTime = std::localtime(&localTime);
startDate = Date(*currentTime);

std::time_t endTime = std::time(nullptr) + duration;
std::tm* timerEndTime = std::localtime(&endTime);
endDate = Date(*timerEndTime);

Your Date constructor should be receiving a std::tm or std::tm const& and store it by value. If you stored a pointer, then all Dates would contain the same pointer to a static internal object.

How do I prevent unique pointers from overlapping in general?

This is easy to do, because std::unique_ptr is more or less fool-proof. As long as you:

• immediately take ownership of pointers to new memory, or
• exclusively use std::make_unique, and
• never call std::unique_ptr::release()

... you won't be able to accidentally make two smart pointers overlap.