为什么在方法签名中添加`mut`会导致借用时间更长？

Question

请考虑以下程序。特殊之处在于，foo和foo_mut的签名要求在生命周期self内借用'a，这也是T<'a>在类型{函数test1<'a>和test2<'a>。

test2可以正常编译，但是test1无法编译。为什么？它们是否具有相同的生命周期语义？我希望这两个功能都将被拒绝，因为t.foo()和t.foo_mut()都应该借用t换来'a，因此直到test1和{{1 }}。

test2

struct T<'a>(&'a u32);

impl<'a> T<'a> {
    fn foo(&'a self) {}
    fn foo_mut(&'a mut self) {}

    fn bar_mut(&mut self) {}
}

fn test1<'a>(mut t: T<'a>) {
    t.foo_mut();
    // The following call is rejected because `t` is still borrowed.
    t.bar_mut();
}

fn test2<'a>(mut t: T<'a>) {
    t.foo();
    // The following call is allowed; it seems that `t` is not borrowed.
    t.bar_mut();
}

fn main() {}

Link to Rust Playground