Rust 锈蚀寿命说明符的问题

考虑以下计划：

fn func(source: &str, state: & Vec<&str>) {
    println!("{}", source);
    println!("{}", state[0]);
}
fn step<'a>(source: &'a str, state: &mut Vec<&'a str>) {
    state.push(&source[4..10]);
    func(source, state);
    state.push(&source[4..10]);
}
fn main() {
    let source = "abcdefghijklmnopqrstuvwxyz";
    {
        let mut state = Vec::<&str>::new();
        step(source, &mut state);
        step(source, &mut state);
        step(source, &mut state);
    }
}

如果有人能告诉我怎么做，我将不胜感激

---

注意：在这个示例程序中，“source”是一个具有静态生存期的字符串。在“真实”程序中，“源”没有静态生存期，而是程序在运行时读取的文件内容。“源”的寿命等于或长于“状态”

函数参数的寿命情况如下：

fn step<'a>(source: &'a str,
            state: &mut Vec<&'a str>)

struct MyStruct<'a, 'b> {
    pub source: &'a str,
    pub state: &'b mut Vec<&'b str>,
}
fn func(arg: MyStruct) {
    println!("{}", arg.source);
    println!("{}", arg.state[0]);
}
fn step<'a>(source: &'a str,
            state: &mut Vec<&'a str>) {
    state.push(&source[4..10]);
    let s = MyStruct{source: source, state: state};
    func(s);
    state.push(&source[4..10]);
}
fn main() {
    let source = "abcdefghijklmnopqrstuvwxyz";
    {
        let mut state = Vec::<&str>::new();
        step(source, &mut state);
        step(source, &mut state);
        step(source, &mut state);
    }
}

107 |             state: &mut Vec<&'a str>) {
    |             ----- consider changing the type of `state` to `&'a mut std::vec::Vec<&'a str>`
108 |     state.push(&source[4..10]);
109 |     let s = MyStruct{source: source, state: state};
    |                                             ^^^^^ lifetime `'a` required

117 |         step(source, &mut state);
    |                           ----- first mutable borrow occurs here
118 |         step(source, &mut state);
    |                           ^^^^^ second mutable borrow occurs here
119 |         step(source, &mut state);
120 |     }
    |     - first borrow ends here

fn step<'a>(source: &'a str,
            state: &mut Vec<&'a str>)

fn step<'a, 'x>(source: &'a str,
                state: &'x mut Vec<&'a str>)

struct MyStruct<'a: 'x, 'x> {
    pub source: &'a str,
    pub state: &'x mut Vec<&'a str>,
}

{
    let s = MyStruct{source: source, state: state};
    func(s);
}