Are there are any side effects of using this method to convert a string to an integer

When you use parseFloat, or parseInt, the conversion is less strict. 1b5 -> 1.

Using 1*number or +number to convert will result in NaN when the input is not valid number. Though unlike parseInt, floating point numbers will be parsed correctly.

Table covering all possible relevant options.

//Variables    // parseInt  parseFloat  + 1* /1   ~~ |0 ^1 >>0  >>>0
var a="123,",//   123        123       NaN       0     & <<0   0
    b = '1.e3',//   1          1000      1000      1000          1000
    c="1.21",//   1          1.21      1.21      1             1
    d = '0020',//   16         20        20        20            20
    e="0x10",//   16         0         16        16            16
    f="3e9", //   3          3000000000  <--    -1294967296    3000000000
    g = '3e10',//   3          30000000000 <--    -64771072      4230196224
    h = 3e25  ,//   3          3e+25     3e+25     0             0
    i = '3e25',//   3          3e+25     3e+25     0             0
    j = 'a123',//   NaN        NaN       NaN       0             0
    k = '  1 ',//   1          1         1         1             1
    l="    ",//   NaN        NaN       0         0             0
    m = '.1  ',//   NaN        0.1       0.1       1             1
    n = '1.  ',//   1          1         1         1             1
    o = '1e999',//  1          Infinity  Infinity  0             0
    p = '1e-999',// 1          0         0         0             0
    q = false ,//   NaN        NaN       0         0             0
    r = void 0,//   NaN        NaN       NaN       0             0
    _ = function(){return 1;}, /* Function _ used below */
    s={valueOf:_},//NaN        NaN       1         1             1
    t={toString:_};// 1        1         1         1             1

// Intervals: (-1e+20, +1e20)  (-∞,+∞)   (-∞,+∞)   (-2³¹,+2³¹)   [0, 2³²)
// In FF9 and Chrome 17, Infinity === Math.pow(2, 1024), approx. 1.7976e+308
// In FF9 and Chrome 17, bitwise operators always return 0 after about ±1e+25

Notes on number conversion methods:

• The number conversion always fail if the first character, after trimming white-space, is not a number.
• parseInt returns an integer representation of the first argument. When the radix (second argument) is omitted, the radix depends on the given input.
  0_ = octal (base-8), 0x_ = hexadecimal (base-16). Default: base-10.
  parseInt ignores any non-digit characters, even if the argument was actually a number: See h, i.
  To avoid unexpected results, always specify the radix, usually 10: parseInt(number, 10).
• parseFloat is the most tolerant converter. It always interpret input as base-10, regardless of the prefix (unlike parseInt). For the exact parsing rules, see here.

  The following methods will always fail to return a meaningful value if the string contains any non-number characters. (valid examples: 1.e+0 .1e-1)

• +n, 1*n, n*1, n/1 and Number(n) are equivalent.
• ~~n, 0|n, n|0, n^1, 1^n, n&n, n<<0 and n>>0 are equivalent. These are signed bitwise operations, and will always return a numeric value (zero instead of NaN).
• n>>>0 is also a bitwise operation, but does not reserve a sign bit. Consequently, only positive numbers can be represented, and the upper bound is 2³² instead of 2³¹.
• When passed an object, parseFloat and parseInt will only look at the .toString() method. The other methods first look for .valueOf(), then .toString(). See q – t.
• NaN, “Not A Number”:
  typeof NaN === 'number'
NaN !== NaN. Because of this awkwardness, use isNaN() to check whether a value is NaN.

When to use which method?

• parseFloat( x ) when you want to get as much numeric results as possible (for a given string).
• parseFloat( (x+'').replace(/^[^0-9.-]+/,'') ) when you want even more numeric results.
• parseInt( x, 10 ) if you want to get integers.
• +x, 1*x .. if you’re only concerned about getting true numeric values of a object, rejecting any invalid numbers (as NaN).
• ~~, 0| .. if you want to always get a numeric result (zero for invalid).
• >>>0 if negative numbers do not exists.
  The last two methods have a limited range. Have a look at the footer of the table.

The shortest way to test whether a given parameter is a real number is explained at this answer:

function isNumber(n) {
    return typeof n == 'number' && !isNaN(n - n);
}