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如何更改Rust中矢量的长度?

[英]How can I change the length of a vector in Rust?

Editor's note: This question predates Rust 1.0 and syntax and methods have changed since then. 编者注:这个问题早于Rust 1.0,语法和方法从那时起就发生了变化。 Some answers account for Rust 1.0. 一些答案解释了Rust 1.0。

I have a function which I would like to have modify a vector in place. 我有一个函数,我想修改一个向量。

fn f(v: &mut Vec<int>) {
    v = Vec::from_elem(10 as uint, 0i);
}

fn main() {
    let mut v: Vec<int> = Vec::new();
    f(&mut v);
}

But this fails to compile. 但这无法编译。 Specifically, I would like to resize v to contain 10 elements of value zero. 具体来说,我想调整v以包含10个零值元素。 What am I doing wrong? 我究竟做错了什么?

Editor's note: This answer predates Rust 1.0 and is no longer necessarily accurate. 编者注:这个答案早于Rust 1.0,不再一定准确。 Other answers still contain valuable information. 其他答案仍包含有价值的信息。

You're looking for the grow method. 你正在寻找grow方法。

let mut vec = vec![1i,2,3];
vec.grow(4, &10);
println!("{}", vec);

Or a combination of grow and clear . 或者是growclear的结合。

You can browse the docs here: http://static.rust-lang.org/doc/master/std/vec/struct.Vec.html 您可以在此处浏览文档: http//static.rust-lang.org/doc/master/std/vec/struct.Vec.html

Use the clear and resize methods. 使用clearresize方法。 (This seems to be the right answer as of Rust 1.0 and 1.5, respectively.) (这似乎是Rust 1.0和1.5的正确答案。)

fn f(v: &mut Vec<u32>) {
    // Remove all values in the vector
    v.clear();

    // Fill the vector with 10 zeros
    v.resize(10, 0);
}

fn main() {
    let mut v: Vec<u32> = Vec::new();
    f(&mut v);
    assert!(v == [0,0,0,0,0,0,0,0,0,0]);
}

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