我想分支并决定在运行时在函数中使用的Trait实现(请参阅下面的代码示例中的poly_read
)。 trait对象是在if表达式的分支臂内部构造的,只需要在poly_read
的生命中存活但我需要Box
它因为该特征不能从表达式中借用手臂,直到我试图分配它的绑定。
我从逻辑上理解为什么借款过早结束,但似乎借用检查器应该能够在if表达式的值绑定时将借用扩展到周围的范围。我意识到这可能是一个天真的想法,但我想更多地了解为什么它不可能。
我对我现在的解决方案有点不满意,因为它需要堆分配,即使我觉得我不需要堆分配,因为我只保留了该功能的生命周期。我想这是因为我们不知道在采用分支之前堆栈上需要的reader
的大小,但是它不能仅仅在编译器中表示为联合,因为我们至少知道最大尺寸。
顺便说一句,我实际上不知道我对所分配的Box
堆的关注是多么有效。一般来说,装箱价值有多贵?
#![feature(io)]
#![feature(path)]
const BYTES: &'static [u8] = &[1u8, 2, 3, 4, 5];
const PATH: &'static str = "/usr/share/dict/words";
use std::old_io::{File, Reader, BufReader};
fn read(r: &mut Reader) {
let some_bytes = r.read_exact(5).unwrap();
assert!(some_bytes.len() == 5);
println!("{:?}", some_bytes);
}
fn poly_read(from_file: bool) {
// Is there any way to extend the lifetime of the ``&mut Reader`` in these branch arms without
// boxing them as I'm doing now. It seems wasteful to do a heap allocation when the actual
// borrow only needs to happen for body of poly_read?
let mut reader = if from_file {
Box::new(File::open(&Path::new(PATH)).unwrap()) as Box<Reader>
// Would like to say:
// File::open(&Path::new(FILE)).unwrap() as &mut Reader
} else {
Box::new(BufReader::new(BYTES)) as Box<Reader>
// Would like to say:
// BufReader::new(BYTES) as &mut Reader
};
// It feels like I'd like the lifetime of values returned from if expressions to be of the
// surrounding scope, rather than the branch arms.
read(&mut reader);
}
fn main() {
poly_read(true);
poly_read(false);
}
答案 0 :(得分:1)
正如@Shepmaster所指出的那样,有一种方法与this answer中的previous question类似。
解决此问题的方法是预先声明两个必要变量:File
和BufReader
:
fn poly_read(from_file: bool) {
// These two variables are predeclared so that they are in scope as
// long as `reader` is
let mut file_reader;
let mut buf_reader;
let mut reader = if from_file {
file_reader = File::open(&Path::new(PATH)).unwrap();
&mut file_reader as &mut Reader
} else {
buf_reader = BufReader::new(BYTES);
&mut buf_reader as &mut Reader
};
read(&mut reader);
}