Reliable JS rounding numbers with toFixed(2) of a 3 decimal number

问题

I am simply trying to round up 1.275.toFixed(2) and I was expecting a return of 1.28, rather than 1.27.

Using various calculators and the simple method of rounding to the nearest hundredth, if the last digit is greater than or equal to five, it should round up.

If this doesn't work with toFixed(2), how would it?

People asking whether console.log(1.275.toFixed(2)) prints off 1.28, here's a quick screenshot MacOS Chrome Version 55.0.2883.95 (64-bit)

回答1:

The toFixed() method is unreliable in its rounding (see Álvaro González' answer as to why this is the case).

In both current Chrome and Firefox, calling toFixed() yields the following inconsistent results:

35.655.toFixed(2) // Yields "36.66" (correct)
35.855.toFixed(2) // Yields "35.85" (wrong, should be "35.86")

MDN describes a reliable rounding implementation:

// Closure
(function() {
  /**
   * Decimal adjustment of a number.
   *
   * @param {String}  type  The type of adjustment.
   * @param {Number}  value The number.
   * @param {Integer} exp   The exponent (the 10 logarithm of the adjustment base).
   * @returns {Number} The adjusted value.
   */
  function decimalAdjust(type, value, exp) {
    // If the exp is undefined or zero...
    if (typeof exp === 'undefined' || +exp === 0) {
      return Math[type](value);
    }
    value = +value;
    exp = +exp;
    // If the value is not a number or the exp is not an integer...
    if (isNaN(value) || !(typeof exp === 'number' && exp % 1 === 0)) {
      return NaN;
    }
    // Shift
    value = value.toString().split('e');
    value = Math[type](+(value[0] + 'e' + (value[1] ? (+value[1] - exp) : -exp)));
    // Shift back
    value = value.toString().split('e');
    return +(value[0] + 'e' + (value[1] ? (+value[1] + exp) : exp));
  }

  // Decimal round
  if (!Math.round10) {
    Math.round10 = function(value, exp) {
      return decimalAdjust('round', value, exp);
    };
  }
  // Decimal floor
  if (!Math.floor10) {
    Math.floor10 = function(value, exp) {
      return decimalAdjust('floor', value, exp);
    };
  }
  // Decimal ceil
  if (!Math.ceil10) {
    Math.ceil10 = function(value, exp) {
      return decimalAdjust('ceil', value, exp);
    };
  }
})();

// Round
Math.round10(55.55, -1);   // 55.6
Math.round10(55.549, -1);  // 55.5
Math.round10(55, 1);       // 60
Math.round10(54.9, 1);     // 50
Math.round10(-55.55, -1);  // -55.5
Math.round10(-55.551, -1); // -55.6
Math.round10(-55, 1);      // -50
Math.round10(-55.1, 1);    // -60
Math.round10(1.005, -2);   // 1.01 -- compare this with Math.round(1.005*100)/100 above
// Floor
Math.floor10(55.59, -1);   // 55.5
Math.floor10(59, 1);       // 50
Math.floor10(-55.51, -1);  // -55.6
Math.floor10(-51, 1);      // -60
// Ceil
Math.ceil10(55.51, -1);    // 55.6
Math.ceil10(51, 1);        // 60
Math.ceil10(-55.59, -1);   // -55.5
Math.ceil10(-59, 1);       // -50

回答2:

The 1.275 base 10 number has finite digits but becomes periodic when converted to base 2:

= 0b1.01000110011001100110011001100110011001100110011010
         ^^^^

Since it has infinite digits, it cannot be represented exactly in a computer unless you use an arbitrary precision library (a library than represents numbers as text to keep them in base 10). JavaScript numbers do not use such library for performance reasons.

Since the original value has already lost precision when it reaches JavaScript, rounding it will not improve that.

回答3:

According to Robby's answer, MDN describes a reliable rounding implementation, therefore I stripped it down to the following snippet to solve my issue of rounding a 3 decimal number of 1.275 to 1.28. Tested in FF4, Chrome 55 and Safari 10.0.3 on MacOS

function decimalAdjust(c,a,b){if("undefined"===typeof b||0===+b)return Math[c](a);a=+a;b=+b;if(isNaN(a)||"number"!==typeof b||0!==b%1)return NaN;a=a.toString().split("e");a=Math[c](+(a[0]+"e"+(a[1]?+a[1]-b:-b)));a=a.toString().split("e");return+(a[0]+"e"+(a[1]?+a[1]+b:b))}Math.round10||(Math.round10=function(c,a){return decimalAdjust("round",c,a)});

Math.round10(1.275, -2);

来源：https://stackoverflow.com/questions/42109818/reliable-js-rounding-numbers-with-tofixed2-of-a-3-decimal-number

标签

javascript

rounding

tofixed