code wars quiz: toInteger

Your task is to program a function which converts any input to an integer.

Do not perform rounding, the fractional part should simply be discarded.

If converting the input to an integer does not make sense (with an object, for instance), the function should return 0 (zero).

Also, Math.floor(), parseInt() and parseFloat() are disabled for your unconvenience.

Onegaishimasu!

function toInteger(n) {
  var arr, i, ret = 0, sgn = 1, mul;
  var pos, front, matches, swap;
  if(typeof n == 'string' || typeof n == 'number'){
    n += '';
    if(n.slice(0, 1) == '-'){
      sgn = -1;
      n = n.slice(1);
    }
    if(matches = /^(\d*\.\d+)e(-?\d+)$/i.exec(n)){
      front = matches[1];
      mul = toInteger(matches[2]);
      pos = front.indexOf('.');
      arr = front.split('');
      if(mul >= 0){
        for(i = 0; i < mul; ++i){
          swap = arr[pos + i];
          arr[pos + i] = arr[pos + i + 1];
          arr[pos + i + 1] = swap;
          if(arr[arr.length - 1] == '.')arr.push('0');
        }
      }else{
        if(pos < -mul){
          for(i = -mul - pos; i > 0; --i){
            arr.unshift('0');
          }
        }
        for(i = 0; i > mul; --i){
          swap = arr[pos + i];
          arr[pos + i] = arr[pos + i - 1];
          arr[pos + i - 1] = swap;
        }
      }
    }else{
      arr = n.split('');
    }
    for(i = 0; i < arr.length; ++i){
      if(arr[i] >= '0' && arr[i] <= '9'){
        ret = ret * 10 + (arr[i] - '0');
      }else if(arr[i] == 'e' || arr[i] == 'E'){
        mul = arr.slice(i + 1).join('');
        if(!/^-?\d+$/.test(mul)){
          ret = 0;
          break;
        }
        mul = toInteger(mul);
        ret *= (mul > 0? Math.pow(10, mul): Math.pow(.1, - mul));
        ret = toInteger(ret);
        break;
      }else if(arr[i] == '.'){
        break;
      }else{
        ret = 0;
        break;
      }
    }
  }
  else if(n === true)return 1;
  return ret * sgn;
}

 

It passes all my unit tests, can not figure out what is wrong with my code.

Test.assertEquals(toInteger(1), 1)
Test.assertEquals(toInteger('1e-2'), 0)
Test.assertEquals(toInteger('-20e-2'), 0)
Test.assertEquals(toInteger('-.98720e-2'), 0)
Test.assertEquals(toInteger('-200.67e-2'), -2)
Test.assertEquals(toInteger('-200.67e3'), -200670)
Test.assertEquals(toInteger('-200.67e0'), -200)
Test.assertEquals(toInteger('-1234e-5'), 0)
Test.assertEquals(toInteger('1234e7'), 12340000000)
Test.assertEquals(toInteger('-200e-2'), -2)
Test.assertEquals(toInteger('e-0'), 1)
Test.assertEquals(toInteger('.3e1'), 3)
Test.assertEquals(toInteger('5000e-2'), 50)
Test.expect(toInteger("4.55") === 4)

 后面我修了一下，简化了这部分的逻辑，加入了对16进制、8进制和单元素数组的处理……最后终于被迫用eval破了这个kata。

看了其余人的答案，发现问题的关键是用位运算符，它自带Int32的操做……

例如~~n， n >> 0， n | 0， n ^ 0，这些都是可行的直接切割整数部分的技巧。