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可以解析为 u128 的最大位数是多少。我Err(ParseIntError { kind: Overflow })在尝试解析 50 位正整数时遇到。

我的错误:

...
...
result: id 89 "40789923115535562561142322423255033685442488917353"      Err(ParseIntError { kind: Overflow })
result: id 90 "44889911501440648020369068063960672322193204149535"      Err(ParseIntError { kind: Overflow })
result: id 91 "41503128880339536053299340368006977710650566631954"      Err(ParseIntError { kind: Overflow })
result: id 92 "81234880673210146739058568557934581403627822703280"      Err(ParseIntError { kind: Overflow })
result: id 93 "82616570773948327592232845941706525094512325230608"      Err(ParseIntError { kind: Overflow })
result: id 94 "22918802058777319719839450180888072429661980811197"      Err(ParseIntError { kind: Overflow })
result: id 95 "77158542502016545090413245809786882778948721859617"      Err(ParseIntError { kind: Overflow })
result: id 96 "72107838435069186155435662884062257473692284509516"      Err(ParseIntError { kind: Overflow })
result: id 97 "20849603980134001723930671666823555245252804609722"      Err(ParseIntError { kind: Overflow })
result: id 98 "53503534226472524250874054075591789781264330331690"      Err(ParseIntError { kind: Overflow })
...
...

对应的代码:

fn read_num(a: &mut Reader<File>) -> Result<u128, Error> {
    let mut sum: u128 = 0;
    for(idx, res) in a.records().enumerate() {
        let res = res.unwrap();
        let val: StringRecord = res;
        let ii = val.get(0).unwrap().trim().parse::<u128>().unwrpa();
        println!("result: id {} {:?}\t {:?}", idx, val.get(0).unwrap(), ii);
    }; // This is formatted
    Ok(sum)
}

我最初认为它必须是由于每行末尾的换行符,但trim应该删除它并且在我看来它正在删除,因为未解析的输出似乎除了\t格式化程序之外没有任何东西println!

溢出是由于固有的限制还是我做错了什么?

4

1 回答 1

2

最大u128的是 2**128-1 = 340282366920938463463374607431768211455,小于 50 位,所以你必须使用struct num::bigint::BigInt(从num板条箱中)。

FromStr您可以通过trait解析大整数。这是一个解析两个 50 位数字并将它们相乘的示例(超级奖励乐趣时间):

extern crate num;

use num::BigInt;
use std::str::FromStr;
use std::ops::Mul;

fn main() {
    let x = BigInt::from_str("9879878782352398572398755757923351299981243778899").unwrap();
    let y = BigInt::from_str("3234235766473868388883432903721391827312463782828").unwrap();
    println!("{}", x.mul(y));
}
于 2019-06-15T15:20:31.073 回答