Understanding std::move and unique_ptr
c++, c++11, move, smart-pointers
Solution
The key concept here is that `std::move` by itself won't do any moving. You can think of it as marking the object as a object that can be moved from.
The signature for `function_call_move` is
void function_call_move( unique_ptr<int>&& ptr );
Which means it can only receive objects that could be moved from, formally known as rvalues, and bind that to a reference. The act of associating an rvalue to a rvalue reference don't invalidate the state of the original object either.
So, unless `function_call_move` actually moves `ptr` to another `std::unique_ptr`inside it, your call to `function_call_move(std::move(intptr));` won't invalidate `intptr` and your usage will be perfectly fine.
Problem
I am new to c++11 and trying to understand to meaning of `std::move` and `unique_ptr` and wrote the following code, which I use `std::move` on a `unique_ptr` in two different ways: ``` void unique_ptr_plain_move() { unique_ptr<int> intptr(new int(10)); unique_ptr<int> intptr2; printf("*intptr = %d\n", *intptr); intptr2 = std::move(intptr); printf("*intptr2 = %d\n", *intptr2); // as expected, crash here as we have already moved intptr's ownership. printf("*intptr = %d\n", *intptr); } ///////////////////////////////////////////// void function_call_move(unique_ptr<int>&& intptr) { printf("[func] *intptr = %d\n", *intptr); } void unique_ptr_function_call_move() { unique_ptr<int> intptr(new int(10)); printf("*intptr = %d\n", *intptr); function_call_move(std::move(intptr)); // this does not crash, intptr still has the ownership of its pointed instance .... printf("*intptr = %d\n", *intptr); } ``` In `unique_ptr_plain_move()`, `intptr2` takes the ownership of `intptr` after `std::move` and therefore we can no longer use `intptr`. However, in `unique_ptr_function_call_move()`, when using `std::move` in a function call, `intptr` still have its ownership of its pointed instance. Can I know what exactly happened when we pass a `std::move(unique_ptr)` to a function? Thank you.