Develop Reference

Javascript: "+" sign concatenates instead of giving sum of variables

I am currently creating a site that will help me quickly answer physics questions.
As it happens, the code didn't run as expected, here is the code
if (option == "dv") {
var Vinitial = prompt("What is the Velocity Initial?")
var acceleration = prompt("what is the acceleration?")
var time = prompt("what is the time?")
Vfinal = Vinitial + acceleration * time
displayV.innerHTML = "v= vf= " + Vfinal + "ms" + sup1.sup();
}
Now, let's say Vinitial was 9, acceleration was 2, and time was 3.
When the code runs, instead of getting 15 for "Vfinal", I get 96.
I figured out that it multiplies acceleration and time fine, and then just concatenates the 9 at the beginning, with 6 (the product of 2 * 3).
I have fixed it for now by using
Vfinal = acceleration * time - (-Vinitial)
which avoids using the "+" sign, but I don't want to have to keep doing this. How do I fix it?

you are dealing with strings here, and math operations on strings will mess up. Remember when ever you are doing math operations you have to convert the data into actual numbers and then perform the math.
Use parseInt() more Details here
Your code should change to
Vfinal = parseInt(Vinitial,10) + parseInt(acceleration,10) * parseInt(time,10);
Edit 1: If the numbers are decimal values then use parseFloat() instead
So the code would be
Vfinal = parseFloat(Vinitial) + parseFloat(acceleration) * parseFloat(time);

Object-Oriented JavaScript - Second Edition: As you already know, when you use the plus sign with two numbers, this
is the arithmetic addition operation. However, if you use the plus
sign with strings, this is a string concatenation operation, and it
returns the two strings glued together:
var s1 = "web";
var s2 = "site";
s1 + s2; // website
The dual purpose of the + operator is a source of errors. Therefore,
if you intend to concatenate strings, it's always best to make sure
that all of the operands are strings. The same applies for addition;
if you intend to add numbers, make sure the operands are numbers.
You can use "+" operator with prompt() to convert returned values from string to int
var Vinitial = +prompt("What is the Velocity Initial?");
var acceleration = +prompt("what is the acceleration?");
var time = +prompt("what is the time?");
Explanation:
var a = prompt('Enter a digit');
typeof a; // "string"
typeof +a; // "number"
If you will enter non-digit data +a gives you NaN. typeof NaN is "number" too :)
You will get the same result with parseInt():
var Vinitial = parseInt(prompt("What is the Velocity Initial?"), 10);
var acceleration = parseInt(prompt("what is the acceleration?"), 10);
var time = parseInt(prompt("what is the time?"), 10);
developer.mozilla.org: parseInt(string, radix);
string: The value to parse.
radix: An integer between 2 and 36 that represents the radix (the base in mathematical numeral systems) of the above mentioned string.
Specify 10 for the decimal numeral system commonly used by humans.
Always specify this parameter to eliminate reader confusion and to
guarantee predictable behavior. Different implementations produce
different results when a radix is not specified, usually defaulting
the value to 10.
Epilogue:
Object-Oriented JavaScript - Second Edition: The safest thing to do is to always specify the radix. If you omit the radix, your code
will probably still work in 99 percent of cases (because most often
you parse decimals), but every once in a while it might cause you a
bit of hair loss while debugging some edge cases. For example, imagine
you have a form field that accepts calendar days or months and the
user types 06 or 08.
Epilogue II:
ECMAScript 5 removes the octal literal values and avoids the confusion
with parseInt() and unspecified radix.

The Problem is, Your value has been took it in a form of string .. so convert your value into Int using parseInt(accelaration).. then it will work ..
Vfinal = parseInt(Vinitial) + parseInt(acceleration) * parseInt(time)

//use ParseInt
var a=10,b=10;
var sum=parseInt(a+b);
ex:
parseInt(Vinitial + acceleration) * time

JS Function "Object expected At line 2 character 0"

I'm building a plan file for XMPie Uplan. Javascript functions are allowed, so as I am learning JS, I thought I would take advantage of them. They seem friendlier than the proprietary QLingo functions. I don't think that should matter though, it is just JS. Anyone see a problem with my JS?
function cents(p) {
var monfor = toString(parseFloat(Math.round(p * 100) / 100).toFixed(2));
return monfor.slice(-2);
}
The purpose of this code is to return just the cents in a price.
Here is what is supposed to be going on:
First I make sure the number has two decimal places and convert to a string. Then I slice off the last two digits of my string leaving me with a 2 digit integer as a string which is the number of pennies in my price. This will flow into the cents portion of a price field with the cents in superscript. (I have another function that uses floor to kill the sub dollar part of the price that populates the dollar part of the price.) The error on this function is:
Error: cents: An error occurred while executing the function script.
Description: Object expected At line 2 character 0.
Thanks in advance for any help!

var numToParse = parseFloat(Math.round(p * 100) / 100).toFixed(2);
var monfor = numToParse.toString();
return monfor.slice(-2);
You were using the two string in the wrong way.
As mentioned in the comments, i didnt know this, you dont even need to parse it to string as the toFixed() parses it itself.
var monfor = parseFloat(Math.round(p * 100) / 100).toFixed(2);
return monfor.slice(-2);

Javascript convert string to integer

I am just dipping my toe into the confusing world of javascript, more out of necessity than desire and I have come across a problem of adding two integers.
1,700.00 + 500.00
returns 1,700.00500.00
So after some research I see that 1,700.00 is being treated as a string and that I need to convert it.
The most relevant pages I read to resolve this were this question and this page. However when I use
parseInt(string, radix)
it returns 1. Am I using the wrong function or the an incorrect radix (being honest I can't get my head around how I decide which radix to use).
var a="1,700.00";
var b=500.00;
parseInt(a, 10);

Basic Answer
The reason parseInt is not working is because of the comma. You could remove the comma using a regex such as:
var num = '1,700.00';
num = num.replace(/\,/g,'');
This will return a string with a number in it. Now you can parseInt. If you do not choose a radix it will default to 10 which was the correct value to use here.
num = parseInt(num);
Do this for each of your string numbers before adding them and everything should work.
More information
How the replace works:
More information on replace at mdn:
`/` - start
`\,` - escaped comma
`/` - end
`g` - search globally
The global search will look for all matches (it would stop after the first match without this)
'' replace the matched sections with an empty string, essentially deleting them.
Regular Expressions
A great tool to test regular expressions: Rubular and more info about them at mdn
If you are looking for a good tutorial here is one.
ParseInt and Rounding, parseFloat
parseInt always rounds to the nearest integer. If you need decimal places there are a couple of tricks you can use. Here is my favorite:
2 places: `num = parseInt(num * 100) / 100;`
3 places: `num = parseInt(num * 1000) / 1000;`
For more information on parseInt look at mdn.
parseFloat could also be used if you do not want rounding. I assumed you did as the title was convert to an integer. A good example of this was written by #fr0zenFry below. He pointed out that parseFloat also does not take a radix so it is always in base10. For more info see mdn.

Try using replace() to replace a , with nothing and then parseFloat() to get the number as float. From the variables in OP, it appears that there may be fractional numbers too, so, parseInt() may not work well in such cases(digits after decimal will be stripped off).
Use regex inside replace() to get rid of each appearance of ,.
var a = parseFloat('1,700.00'.replace(/,/g, ''));
var b = parseFloat('500.00'.replace(/,/g, ''));
var sum = a+b;
This should give you correct result even if your number is fractional like 1,700.55.
If I go by the title of your question, you need an integer. For this you can use parseInt(string, radix). It works without a radix but it is always a good idea to specify this because you never know how browsers may behave(for example, see comment #Royi Namir). This function will round off the string to nearest integer value.
var a = parseInt('1,700.00'.replace(/,/g, ''), 10); //radix 10 will return base10 value
var b = parseInt('500.00'.replace(/,/g, ''), 10);
var sum = a+b;
Note that a radix is not required in parseFloat(), it will always return a decimal/base10 value. Also, it will it will strip off any extra zeroes at the end after decimal point(ex: 17500.50 becomes 17500.5 and 17500.00 becomes 17500). If you need to get 2 decimal places always, append another function toFixed(decimal places).
var a = parseFloat('1,700.00'.replace(/,/g, ''));
var b = parseFloat('500.00'.replace(/,/g, ''));
var sum = (a+b).toFixed(2); //change argument in toFixed() as you need
// 2200.00
Another alternative to this was given by #EpiphanyMachine which will need you to multiply and then later divide every value by 100. This may become a problem if you want to change decimal places in future, you will have to change multiplication/division factor for every variable. With toFixed(), you just change the argument. But remember that toFixed() changes the number back to string unlike #EpiphanyMachine solution. So you will be your own judge.

try this :
parseFloat(a.replace(/,/g, ''));
it will work also on : 1,800,300.33
Example :
parseFloat('1,700,800.010'.replace(/,/g, '')) //1700800.01

Javascript doesn't understand that comma. Remove it like this:
a.replace(',', '')
Once you've gotten rid of the comma, the string should be parsed with no problem.

Javascript is treating variables as Strings, why?

I have the variable y, which is a subtotal. Its value is different depending on what happens with the html, but throughout the script I declared it like this:
var y = 21.78;
etc. Why is it that on my last equation where I add up the total, it treats them as strings when I want to add the values?
var tax = (0.055*y).toFixed(2);
var totalprice = y+tax;
/* totalprice holds "21.781.20" instead of 22.98 */

According to:
https://developer.mozilla.org/en/JavaScript/Reference/Global_Objects/Number/toFixed
toFixed() returns:
A string representation of number that does not use exponential
notation and has exactly digits digits after the decimal place.
thus, y+tax is cast to a string since one of the operands is a string.
In my opinion, this would make sense as Javascript's intrinsic numeric types do not have the ability to store a specific number of decimal place digits, so a string would be the most appropriate data structure to store this with.
I would advise you do all your addition before calling toFixed(), since the method is most suitable for formatting display output.
var taxRate = 0.055;
var subtotal = 21.78;
var tax = (taxRate * subtotal).toFixed(2),
totalprice = ((1+taxRate) * subtotal).toFixed(2);
document.write(totalprice);

The .toFixed() method returns a string. Try applying that method as the last step after all other calculations.

Here's a simple fix. Put '+' in front of the tax variable to convert it to a number.
var y = 21.78;
var tax = (0.055*y).toFixed(2);
var totalprice = y+ (+tax);
totalprice === 22.98;
If you don't want any rounding errors when you use toFixed, then include this re-implementation of it in your script.
http://bateru.com/news/2012/03/reimplementation-of-number-prototype-tofixed/

In my experience, if there's any chance available, Javascript will see the "+" sign as concatenate rather than addition. It's driven me nuts on more than one occasion. I will generally do this rather than chance concatenation:
var totalprice = parseInt(y)+parseInt(tax);

When letter replaces value, multiply with 1 when you're in need of +.
var totalprice = (y*1) + tax .
Other operands work fine, it's just the + operand that needs special treatment when variable replace value.

How to make 5509.099999999999 as 5509.09 using javascript

How to make 5509.099999999999 as 5509.09 using javascript.

Lots of mathy options that end up with .1 so how about;
var f = 5509.099999999999
if ((f = f.toString()).indexOf(".") >= 0)
f = f.substr(0, 3 + f.indexOf("."))
print(parseFloat(f))
>>5509.09

Have you tried this?
var value = 5509.099999999999;
var str = value.toString();
var result = str.substr(0,7);
Then if you need it to be a float again you can do:
var FinalAnswer = parseFloat(result);
You don't need all these variables, but that is the step by step.

var result = (Math.round((5509.09999 * 100) - 1)) / 100;

You could use .toFixed(2) but this will round the value, so in your example you'll end up with 5509.10 instead of 5509.09.
The next best option is to use Math.floor(), which truncates rather than rounding. Unfortunately, this only gives integer results, so to get the result to 2 decimal places, you'd need to multiply by 100, then use Math.floor(), and then divide by 100 again.
var value = 5509.099999999999;
var result = Math.floor(value*100)/100;
[EDIT]
Hmm, unfortunately, the above doesn't work due to problems with floating point precision -- even just the first step of multiplying it by 100 gives 550910.
Which means that the best answer is likely to be converting it to a string and chopping the string into bits.
var value = 5509.099999999999;
var str_value = value.toString();
var bits = str_value.split('.');
var result = bits[0]+"."+bits[1].substr(0,2);
I wouldn't normally suggest doing string manipulation for this sort of thing, because it is obviously a maths problem, but given the specific requirements of the question, it does seem that this is the only workable solution in this case.

You can truncate the number to a certain number of decimal places using this function:
function truncateNumber(number, digits){
var divisor = Math.pow(10,digits);
return Math.floor(number*divisor)/divisor;
}
If you want to round the number instead, you can use JavaScript's built in Number.toFixed function. If you always want the number a certain number of digits long, you can use the Number.toPrecision function.

if you want to take two decimal places, you can use .toPrecision(n) javascript function, where n is the total number of digits desired.
so, for your example, you'd have to do
var x = 5509.099999999999;
x = x.toPrecision(6);
this, however, rounds results in 5509.10