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

Question

I'm just trying to round 1.275.toFixed(2), and I was expecting 1.28 to return , not 1.27.

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

If this does not work with toFixed (2), how is it?

People ask if console.log (1.275.toFixed (2)) will print 1.28, here's a quick screenshot of the MacOS Chrome Version 55.0.2883.95 (64-bit)

Answer 1

toFixed() (. Álvaro González ', ).

Chrome Firefox toFixed() :

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

MDN :

// 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

Answer 2

10 1.275 , 2 :

= 0b1.01000110011001100110011001100110011001100110011010
         ^^^^

, , (, , 10). JavaScript ​​ .

JavaScript, .

Answer 3

According to Robbie's answer, MDN describes a robust rounding implementation , so I divided it into the following snippet to solve my rounding problem 3 decimal from 1.275to 1.28. Tested on 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);