Does unique_ptr:: release () call the destructor?

The unique_ptr::release() function does not call the destructor of the object it manages. Its primary purpose is to release ownership of the raw pointer managed by the unique_ptr object, effectively disconnecting the unique_ptr from the resource it manages. After calling release(), the unique_ptr becomes a null pointer (no longer managing any resource), and the raw pointer is returned to the caller. Since the unique_ptr no longer manages the raw pointer, the object pointed to by the raw pointer is not automatically released or destroyed, meaning the responsibility for managing the resource transfers to the caller.

For example:

cpp
#include <iostream>
#include <memory>

class Test {
public:
    Test() { std::cout << "Test Constructor\n"; }
    ~Test() { std::cout << "Test Destructor\n"; }
};

int main() {
    std::unique_ptr<Test> ptr = std::make_unique<Test>();
    // Output:
    // Test Constructor

    Test* raw_ptr = ptr.release();
    // At this point, ptr is empty and no longer manages the Test object
    // The Test object is not destroyed, so the destructor is not called

    // Must manually delete the object pointed to by raw_ptr to prevent memory leaks
    delete raw_ptr;
    // Output:
    // Test Destructor

    return 0;
}

In this example, after calling ptr.release(), the unique_ptr relinquishes ownership of the Test object, and the object is not destroyed until we manually call delete. If forgotten, this can lead to a memory leak. This demonstrates that the release() function itself does not trigger the destructor.