CODEWITHSUNDEEP
javascriptmath
PHP Worm...
PHP Worm...

Reputation: 4224

toFixed function not working properly ( please give a reason not an alternative)

toFixed() function responding differently for float values. For Example:

 var a = 2.555;
 var b = 5.555;

console.log(a.toFixed(2));  /* output is 2.56 */ 
console.log(b.toFixed(2));  /* output is 5.55 */

For 2.555/3.555 results are (2.56/3.56)

and

For other values(not sure for all values) it is showing #.55 (# refers to any number)

I am confused can any one help me out.

Thanks in advance.

Upvotes: 0

Views: 3199

Answers (3)

6502
6502

Reputation: 114461

Javascript uses a binary floating point representation for numbers (IEEE754). This format is able to represent exactly (without approximation) only numbers that can be expressed in the form n/2m where both n and m are integers.

Any number that is not a rational where the denominator is an integral power of two cannot be represented exactly because in binary it is a periodic number (it has infinite binary digits after the point).

The number 0.5 (i.e. 1/2) is fine, (in binary is just 0.1₂) but for example 0.55 (i.e. 11/20) cannot be represented exactly (in binary it's 0.100011001100110011₂… i.e. 0.10(0011)₂ with the last part 0011₂ repeating infinite times).

If you need to do any computation in which the result depends on exact decimal numbers you need to use an exact decimal representation. A simple solution if the number of decimals is fixed (e.g. 3) is to keep all values as integers by multiplying them by 1000...

2.555 --> 2555
5.555 --> 5555
3.7   --> 3700

and adjusting your computation when doing multiplications and divisions accordingly (e.g. after multiplying two numbers you need to divide the result by 1000).

The IEEE754 double-precision format is accurate with integers up to 9,007,199,254,740,992 and this is often enough for prices/values (where the rounding is most often an issue).

Upvotes: 8

A.B
A.B

Reputation: 20445

toFixed() method depending on Browser rounds down or retain.

Here is the solution for this problem, check for "5" at the end 

 var num = 5.555;
 var temp = num.toString();
if(temp .charAt(temp .length-1)==="5"){
temp = temp .slice(0,temp .length-1) + '6';
}
num = Number(temp);
Final = num.toFixed(2);

Or reusable function would be like

function toFixedCustom(num,upto){

     var temp = num.toString();
    if(temp .charAt(temp .length-1)==="5"){
    temp = temp .slice(0,temp .length-1) + '6';
    }
    num = Number(temp);
    Final = num.toFixed(upto);
    return Final;
}

var a = 2.555;
 var b = 5.555;

console.log(toFixedCustom(a,2));  
console.log(toFixedCustom(b,2));

Upvotes: 0

I'm Geeker
I'm Geeker

Reputation: 4637

Try this Demo Here

function roundToTwo(num) {    
    alert(+(Math.round(num + "e+2")  + "e-2"));
}

roundToTwo(2.555);
roundToTwo(5.555);

Upvotes: 3

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