Problem of the Day
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Continuing on with the theme of numbers, can you reimplement a basic implementation of parseInt? Our function will take in a string and return an integer representation of that string if one exists. Otherwise you can throw an error or return not a number. It's safe to assume that the input will be of base 10 but for bonus points I encourage you to support multiple bases as does the real parseInt.

parseInt('444') -> 444
parseInt('J3110') -> throw error or return NaN

Comments:

• Anonymous - 10 years, 1 month ago

  def parseInt(st):
      res = 0
      for ch in st:
          if not '0' <= ch <= '9':
              raise ValueError
          res *= 10
          res += ord(ch)-ord('0')
      return res
  

  Only works for base 10 so no bonus points

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• Anonymous - 10 years, 1 month ago

  C++:

  #include <string>
  #include <iostream>
  #include <cmath>
  
  int parseInt(std::string s) {
      int answer = 0;
      for(unsigned int i = 0; i < s.size(); i++) {
          if((int)s[i] < 48 || (int)s[i] > 57) {
              return 0;
          }
          answer += (((int)s[i] - 48) * pow(10, s.size() - 1 - i));
      }
      return answer;
  }
  
  int main(int argc, char* argv[]) {
      std::cout << parseInt("444") << std::endl;
      std::cout << parseInt("J3110") << std::endl;
  }
  

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• asheehan - 10 years, 1 month ago

  in Ruby:

  def parse_int(x)
      result = 0
      x = x.lstrip.rstrip
      x.each_byte.with_index do |n,i|
          n -= '0'.unpack('c')[0]
          if n >= 0 and n <= 9
              result += n * 10 ** (x.length - (i+1))
          else # no more numbers, time to return
              if result > 0 or x[0] == '0'.unpack('c')[0]
                  return result / 10 ** (x.length - i)
              else
                  return 'NaN'
              end
          end
      end
  
      return result;
  end
  
  p parse_int('444');
  

  can replace the #unpack with #ord on newer versions of Ruby.

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• bumbleguppy - 10 years, 1 month ago

  Although deprecated, I will allow octal notation

  function myParseInt(num, radix) {
    var digits = ''; 
    var result = 0;
    var i = 0;
    var decplace = 1;
    var idx = 0;
  
    if (typeof radix === 'undefined') { //optional radix arg
      radix = 10;
    } 
  
    //test first 'num' argument
    if (typeof num !== 'string') { 
      if(typeof num === 'number') {
        if (radix !== 10) {
          num = num.toString(); //maybe someone wants to turn a decimal to binary? Okay...
        } else {
          return num;
        }
      } else {
        return NaN;
      }
    } else {
      num = num.toUpperCase().trim();
    }
  
    //test second 'radix' argument
    if (typeof radix !== 'number' || radix > 36 || radix < 2) { 
      throw new Error('myParseInt : Radix argument only takes integer value from 2 to 36');
    } else{
  
      if (num[0] === '0' && radix === 10) {
        if (num[1] === 'X') {
          radix = 16;
          num = num.substr(2);  //hexadecimal notation
        } else {
          radix = 8;
          num = num.substr(1); //ocatal notation
        }
      }
    }
  
    //chop string that will be used to match the characters in the first argument to the length 'radix'
    digits = '0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ'.substr(0, radix);
  
    i = num.length;
  
    while(i--) {
      idx = digits.indexOf(num[i]);
      if (idx === -1) {
        return NaN;
      } else {
        result += (decplace * idx);
        decplace *= radix;
      }
    }
  
    return result;
  }
  console.log(myParseInt('0xFF'));
  

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