面试题：Rust字符串实用函数在复杂场景下的运用

Question

假设有一个字符串`let s = "1,2,3,4,5"`. 编写代码，首先使用`split`函数分割字符串，然后将分割后的子字符串转换为`i32`类型，并计算它们的总和，最后打印出总和。请处理可能出现的转换错误。

Answer 1

fn main() {
    let s = "1,2,3,4,5";
    let parts: Vec<&str> = s.split(',').collect();
    let mut sum = 0;
    for part in parts {
        match part.parse::<i32>() {
            Ok(num) => sum += num,
            Err(_) => println!("转换错误: {}", part),
        }
    }
    println!("总和: {}", sum);
}

s = "1,2,3,4,5"
parts = s.split(',')
sum_value = 0
for part in parts:
    try:
        num = int(part)
        sum_value += num
    except ValueError:
        print(f"转换错误: {part}")
print(f"总和: {sum_value}")

let s = "1,2,3,4,5";
let parts = s.split(',');
let sum = 0;
for (let part of parts) {
    let num = parseInt(part);
    if (isNaN(num)) {
        console.log(`转换错误: ${part}`);
    } else {
        sum += num;
    }
}
console.log(`总和: ${sum}`);

以上分别给出了Rust、Python、JavaScript三种语言实现该功能的代码示例，每种语言都按要求先分割字符串，处理转换错误并计算总和后打印。