Does unique_ptr::release() call the destructor?

Question

Is this code correct?

auto v =  make_unique(12);
v.release();     // is this possible?

Is it equivalent to delete of a

Answer 1

release will leak your raw pointer since you don't assign it to anything.

It is meant to be used for something like

int* x = v.release();

Which means v is no longer managing the lifetime of that pointer, it is delegating the raw pointer ownership to x. If you just release without assigning to anything, you leak the raw pointer.