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c859550e6c
👍 👍 👍 Connects to #6
54 lines
1.3 KiB
Rust
54 lines
1.3 KiB
Rust
// Make me compile without changing line 9! Scroll down for hints :)
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pub fn main() {
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let vec0 = Vec::new();
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let mut vec1 = fill_vec(vec0);
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// Do not change the following line!
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println!("{} has length {} content `{:?}`", "vec0", vec0.len(), vec0);
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vec1.push(88);
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println!("{} has length {} content `{:?}`", "vec1", vec1.len(), vec1);
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}
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fn fill_vec(vec: Vec<i32>) -> Vec<i32> {
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let mut vec = vec;
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vec.push(22);
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vec.push(44);
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vec.push(66);
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vec
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}
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// So `vec0` is being *moved* into the function `fill_vec` when we call it on
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// line 6, which means it gets dropped at the end of `fill_vec`, which means we
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// can't use `vec0` again on line 9 (or anywhere else in `main` after the
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// `fill_vec` call for that matter). We could fix this in a few ways, try them
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// all!
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// 1. 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 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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