assert_eq
宏代码的一部分是:
($left:expr, $right:expr, $($arg:tt)+) => ({
match (&($left), &($right)) {
(left_val, right_val) => {
if !(*left_val == *right_val) {
// The reborrows below are intentional. Without them, the stack slot for the
// borrow is initialized even before the values are compared, leading to a
// noticeable slow down.
$crate::panic!(r#"assertion failed: `(left == right)`
left: `{:?}`,
right: `{:?}`: {}"#, &*left_val, &*right_val,
$crate::format_args!($($arg)+))
}
}
}
});
评论说重借是有意的,它将以某种方式影响堆栈的使用方式。什么是重新借用,它如何影响编译器生成的代码?
最佳答案
我不确定它如何影响编译,但是重新借用是当您取消引用某些内容然后直接借用它时。
例如:
fn main() {
let mut vec: Vec<u8> = vec![1, 2, 3];
let mut_ref: &mut Vec<u8> = &mut vec;
let reborrow: &Vec<u8> = &*mut_ref;
}
它可以用于:缩短生命周期的示例用法(非常人为,但这过于简单了):
fn main() {
let mut outer_vec = vec![1, 2, 3];
let outer_ref = &mut outer_vec;
{
// shorten lifetime of outer_ref
// not doing this causes an error
let mut outer_ref = &mut *outer_ref; // mutable reborrow
let mut inner_vec = vec![1, 2, 3];
let inner_ref = &mut inner_vec;
// imagine a real condition
if true {
outer_ref = inner_ref;
}
// use outer_ref which could point to the outer_vec or inner_vec
println!("{:?}", outer_ref);
}
println!("{:?}", outer_ref);
}
获取对智能指针后面的内容的引用(在这种情况下为Box
):fn main() {
let vec = vec![1, 2, 3];
let boxed = Box::new(vec);
let reborrow: &Vec<u8> = &*boxed;
println!("{:?}", reborrow);
}
指针引用:fn main() {
let num: u8 = 10;
let pointer: *const u8 = &num as *const u8;
unsafe {
let ref_from_ptr: &u8 = &*pointer;
}
}
不确定是否将所有这些严格地视为“借入”。
关于rust - 什么是重新借用,它如何影响编译器生成的代码?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/66288291/