为什么这个Rust比我类似的Python慢​​？

Question

我有以下Rust程序(rustc 1.0.0-nightly(44a287e6e 2015-01-08 17:03:40 -0800)):

use std::io::BufferedReader;
use std::io::File;

fn main() {
    let path = Path::new("nc.txt");
    let mut file = BufferedReader::new(File::open(&path));
    let lines: Vec<String> = file.lines().map(|x| x.unwrap()).collect();
    println!("{}", lines[500]);
}

根据http://doc.rust-lang.org/std/io/上的示例,上面是将文件的行拉成字符串向量的方法.我已经投入了第500行的输出.

为了解决Python中的相同任务,我写了以下内容:

#!/usr/local/bin/python3

def main():
    with open('nc.txt', 'r') as nc:
        lines = nc.read().split('\n')
    print("{}".format(lines[500]))

if __name__ == '__main__':
    main()

当我运行编译的Rust并计时时,我得到了这个:

rts@testbed $ time ./test
A declaration of independence by Kosovo will likely bring a similar declaration from Georgia's breakaway Abkhazia region, which Russia could well recognize.

./test  1.09s user 0.02s system 99% cpu 1.120 total

运行Python给出:

rts@testbed $ time ./test.py 
A declaration of independence by Kosovo will likely bring a similar declaration from Georgia's breakaway Abkhazia region, which Russia could well recognize.
./test.py  0.05s user 0.03s system 90% cpu 0.092 total

我知道这println!是一个扩展到更复杂的宏

::std::io::stdio::println_args(::std::fmt::Arguments::new({
    #[inline]
    #[allow(dead_code)]
    static __STATIC_FMTSTR: &'static [&'static str] = &[""];
    __STATIC_FMTSTR
},
&match (&lines[500],) {
    (__arg0,) => [::std::fmt::argument(::std::fmt::String::fmt, __arg0)],
}));

尽管如此,这似乎不会导致超过一秒的额外执行时间.事实上,这些代码片段是否相似？我是否误解了将线条读入矢量并输出其中一个的最有效方法？

供参考,nc.txt具有以下属性:

rts@testbed $ du -hs nc.txt 
7.5M    nc.txt
rts@testbed $ wc -l nc.txt
   60219 nc.txt

Answer 1

使用优化标志构建它.

cargo run --release