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@ -287,23 +287,24 @@ Also: Try accessing `vec0` after having called `fill_vec()`. See what happens!""
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[[exercises]]
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name = "move_semantics2"
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path = "exercises/move_semantics/move_semantics2.rs"
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mode = "compile"
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mode = "test"
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hint = """
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So, `vec0` is passed into the `fill_vec` function as an argument. In Rust,
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when an argument is passed to a function and it's not explicitly returned,
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you can't use the original variable anymore. We call this "moving" a variable.
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Variables that are moved into a function (or block scope) and aren't explicitly
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returned get "dropped" at the end of that function. This is also what happens here.
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There's a few ways to fix this, try them all if you want:
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1. Make another, separate version of the data that's in `vec0` and pass that
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When running this exercise for the first time, you'll notice an error about
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"borrow of moved value". In Rust, when an argument is passed to a function and
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it's not explicitly returned, you can't use the original variable anymore.
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We call this "moving" a variable. When we pass `vec0` into `fill_vec`, it's being
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"moved" into `vec1`, meaning we can't access `vec0` anymore after the fact.
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Rust provides a couple of different ways to mitigate this issue, feel free to try them all:
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1. You could make another, separate version of the data that's in `vec0` and pass that
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to `fill_vec` instead.
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2. Make `fill_vec` borrow its argument instead of taking ownership of it,
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and then copy the data within the function in order to return an owned
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`Vec<i32>`
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3. Make `fill_vec` *mutably* borrow a reference to its argument (which will need to be
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mutable), modify it directly, then not return anything. Then you can get rid
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of `vec1` entirely -- note that this will change what gets printed by the
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first `println!`"""
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and then copy the data within the function (`vec.clone()`) in order to return an owned
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`Vec<i32>`.
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3. Or, you could make `fill_vec` *mutably* borrow a reference to its argument (which will need to be
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mutable), modify it directly, then not return anything. This means that `vec0` will change over the
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course of the function, and makes `vec1` redundant (make sure to change the parameters of the `println!`
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statements if you go this route)
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"""
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[[exercises]]
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name = "move_semantics3"
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