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Comment by bszupnick | Hacker News Reader

This code doesn't duplicate it. In Rust when a variable is sent as an argument to a function it's "ownership" moves to be in the scope of that function.

https://doc.rust-lang.org/book/ch04-01-what-is-ownership.htm...

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You're missing my point. Unless the only thing you want to do with your giant data structure is measure its size, you're not going to be passing ownership of your only copy of it into the get_size function. You're going to be passing in a copy -- hence the cost of duplicating everything.

In the different contrived case where it gets copied, you'd instead change this code to take an immutable reference to it, and compute the size of that. Or you'd call .size() instead of calling this function!

It is just an example. You can think of "measuring size" here as getting the result of a long computation that involves a lot of allocations. After you get the result, you no longer care about the intermediate stuff – i.e. all the allocations. You certainly don't want to duplicate them, you just want to get rid of them, and the author tells that you might not want to deallocate (drop) them in the UI thread.

If it helps you, you might want to imagine the contents of the get_size function as being the end part of a longer calc_foo function. What's really missing the point is focusing so hard on the part that the example even contains a call to size() of a collection.

Rust is stricter about aliasing than C++ is.

Vectors are the size of 3 pointers (data, size, capacity), so I guess 24 bytes on x64.

Even if the move requires a memcpy, it's only copying that 24 bytes - The heap allocation is not copied, because there are never two owners of the vector at once.

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