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  1. Prepare
  2. Tutorials
  3. 30 Days of Code
  4. Day 16: Exceptions - String to Integer
  5. Discussions

Day 16: Exceptions - String to Integer

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994 Discussions

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  • mirovlab
     + 0 comments

    🐍 If your try: except code can't get validation - try to delete 'if name == 'main':'

    Probably this if statement triggers an error message reading it as part of "if (...) else") statement.

    Graditudes:

    https://www.hackerrank.com/karinaxfarias https://www.hackerrank.com/przemyslaw_siek1

  • adadasaivenkat01
     + 0 comments
    s=input()
try:
    s=int(s)
    print(s)
except ValueError: 
    print("Bad String")

  • andrecezu
     + 1 comment

    I made a code in python using try and except, however this error is appearing in the submition "Error reading result file.You should use exception handling concepts." if name == 'main': S = input() try: integer=int(S) print(str(integer)) except ValueError: print("Bad String") pass except Exception: print("Bad String")

        pass

  • jonlle19
     + 0 comments

    Solution in JavaScript/TypeScript

    In JavaScript/TypeScript, the situation is a bit different. The parseInt function does not throw an exception when the conversion fails. Instead, it returns NaN (Not-a-Number), which means we need additional validation to handle this case. Due to the challenge's restrictions, which do not allow the use of conditionals, it was not possible to submit a solution that met all the requirements. Here is an attempt at a solution in TypeScript using parseInt for conversion, but it requires additional validation for NaN:

    function throwBadString(): never {
    throw new Error("Bad String");
}   

function main() {
    const S: string = readLine();
    
    try {
        const num = parseInt(S);
        isNaN(num) && throwBadString();
        console.log(num);
    } catch (error) {
        console.log("Bad String");
    }
}

    Solution in Java

    In Java, the solution was straightforward thanks to Integer.parseInt, which throws a NumberFormatException if the string cannot be converted to an integer. This allows us to handle the conversion error clearly and simply using a try-catch block. Here's the code:

    public class Solution {
    public static void main(String[] args) throws IOException {
        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));

        String S = bufferedReader.readLine();
        
        try {
            int result = Integer.parseInt(S);
            System.out.println(result);
        } catch (NumberFormatException e) {
            System.out.println("Bad String");
        }

        bufferedReader.close();
    }
}

  • siegfried_bozza
     + 1 comment

    Hi Can someone explain why this makes the test to fail (while the output is ok..) ? Thanks in advance

    'use strict';

process.stdin.resume();
process.stdin.setEncoding('utf-8');

let inputString = '';
let currentLine = 0;

process.stdin.on('data', function(inputStdin) {
    inputString += inputStdin;
});

process.stdin.on('end', function() {
    inputString = inputString.split('\n');

    main();
});

function readLine() {
    return inputString[currentLine++];
}



function main() {
    const S = readLine();
    
     try {
        const parsed = parseInt(S, 10)
        if (!/^\d+$/.test(S) || isNaN(parsed)) {
            throw new Error("Bad String");
        }
        console.log(parsed);
    } catch (err) {
        console.log("Bad String");
    }
}