Rust's moving and copying semantics are very different from C++. I'm going to take a different approach to explain them than the existing answer.
In C++, copying is an operation that can be arbitrarily complex, due to custom copy constructors. Rust doesn't want custom semantics of simple assignment or argument passing, and so takes a different approach.
First, an assignment or argument passing in Rust is always just a simple memory copy.
let foo = bar; // copies the bytes of bar to the location of foo (might be elided)
function(foo); // copies the bytes of foo to the parameter location (might be elided)
But what if the object controls some resources? Let's say we are dealing with a simple smart pointer, Box.
let b1 = Box::new(42);
let b2 = b1;
At this point, if just the bytes are copied over, wouldn't the destructor (drop in Rust) be called for each object, thus freeing the same pointer twice and causing undefined behavior?
The answer is that Rust moves by default. This means that it copies the bytes to the new location, and the old object is then gone. It is a compile error to access b1 after the second line above. And the destructor is not called for it. The value was moved to b2, and b1 might as well not exist anymore.
This is how move semantics work in Rust. The bytes are copied over, and the old object is gone.
In some discussions about C++'s move semantics, Rust's way was called "destructive move". There have been proposals to add the "move destructor" or something similar to C++ so that it can have the same semantics. But move semantics as they are implemented in C++ don't do this. The old object is left behind, and its destructor is still called. Therefore, you need a move constructor to deal with the custom logic required by the move operation. Moving is just a specialized constructor/assignment operator that is expected to behave in a certain way.
So by default, Rust's assignment moves the object, making the old location invalid. But many types (integers, floating points, shared references) have semantics where copying the bytes is a perfectly valid way of creating a real copy, with no need to ignore the old object. Such types should implement the Copy trait, which can be derived by the compiler automatically.
#[derive(Clone, Copy)]
struct JustTwoInts {
one: i32,
two: i32,
}
This signals the compiler that assignment and argument passing do not invalidate the old object:
let j1 = JustTwoInts { one: 1, two: 2 };
let j2 = j1;
println!("Still allowed: {}", j1.one);
Note that trivial copying and the need for destruction are mutually exclusive; a type that is Copy cannot also be Drop.
Now what about when you want to make a copy of something where just copying the bytes isn't enough, e.g. a vector? There is no language feature for this; technically, the type just needs a function that returns a new object that was created the right way. But by convention this is achieved by implementing the Clone trait and its clone function. In fact, the compiler supports automatic derivation of Clone too, where it simply clones every field.
#[derive(Clone)]
struct JustTwoVecs {
one: Vec<i32>,
two: Vec<i32>,
}
let j1 = JustTwoVecs { one: vec![1], two: vec![2, 2] };
let j2 = j1.clone();
And whenever you derive Copy, you should also derive Clone, because containers like Vec use it internally when they are cloned themselves.
#[derive(Copy, Clone)]
struct JustTwoInts { /* as before */ }
Now, are there any downsides to this? Yes, in fact there is one rather big downside: because moving an object to another memory location is just done by copying bytes, and no custom logic, a type cannot have references into itself. In fact, Rust's lifetime system makes it impossible to construct such types safely.
But in my opinion, the trade-off is worth it.
