How to compare two 20 digit numbers in javascript

  • Thread starter balakrishnan.dinesh
  • Start date
B

balakrishnan.dinesh

hi frnds,
Im having two 20digit numbers, But while comparing those it
is giiving wrong ouput in javascript.

for example here is my code,
my secanrio is ,

~ If first 20 digit number is greater number than second 20 digit
number ,then it should return.

* And another important thing is, the two 20digit number will be same
but last one digit only change

Code:

<HTML>
<script language='javascript'>
function numberCheck()
{
var val1=document.getElementById('txt1').value;
var val2=document.getElementById('txt2').value;
if(parseInt(val1) > parseInt(val2))
alert("Greater")
else
alert("Correct");

}
</script>
<body>
<input type=text id="txt1" value="18446744073709551617" size=25>
<input type=text id="txt2" value="18446744073709551616" size=25>
<input type=button onclick="numberCheck();" value="check">

</body>
</html>

So here txt1 value is greater than txt2 value, but it returning Correct
alert.
Please tel me reason why its returning like this,
Is there any restriction for comparing more than 20 digits
 
R

Randy Webb

(e-mail address removed) said the following on 7/28/2006 1:44 AM:
hi frnds,
Im having two 20digit numbers, But while comparing those it
is giiving wrong ouput in javascript.

for example here is my code,
my secanrio is ,

~ If first 20 digit number is greater number than second 20 digit
number ,then it should return.

* And another important thing is, the two 20digit number will be same
but last one digit only change
Click to expand...

If the only thing that can change is the last digit then simply compare
the last character and it doesn't matter how long the "number" is if it
is stored in a string.
<HTML>
<script language='javascript'>
function numberCheck()
{
var val1=document.getElementById('txt1').value;
Click to expand...

var lastChar1 = val1.charAt(val1.length-1);
var val2=document.getElementById('txt2').value;
Click to expand...

var lastChar2 = val2.charAt(val2.length-1);


Now, simply compare the last 2 characters.
if(parseInt(val1) > parseInt(val2))
Click to expand...

Don't use parseInt without the second parameter or you may eventually
get unexpected results.

if (+lastChar1>+lastChar2)

So here txt1 value is greater than txt2 value, but it returning Correct
alert.
Please tel me reason why its returning like this,
Click to expand...

The incorrectness introduced by floating point math.
Is there any restriction for comparing more than 20 digits
Click to expand...

Probably.
 
R

RobG

So here txt1 value is greater than txt2 value, but it returning Correct
alert.
Please tel me reason why its returning like this,
Is there any restriction for comparing more than 20 digits
Click to expand...

Yes, absolutely.

If only the last digit needs to be compared, then Randy has answered
your question. However, if you want to compare large numbers as
numbers, read the following thread:

<URL:
http://groups.google.com/group/comp...gest+positive+integer&rnum=3#8c73213f5f2e91c5
From that thread:
Click to expand...

"The representable range is +/-1.7976931348623157×10^308 (with
numbers as small as +/-5×10^-324). The range of precise integer
values is +/-9007199254740992 (+/-2^53)."

Note that there are many integers in the representable range that can't
be represented exactly, starting with integers of 16 digits. That is
well short of your requirement for 20 digits.
 
B

balakrishnan.dinesh

Hi Randy,

For a example only ,i told u the last character, It can be any digit in
the whole number, So i want to compare the whole 20 digit number with
other 20 digit number.

I want know ,why is it not comparing correctly. Is there anyother way
to compare
 
B

balakrishnan.dinesh

Hi Randy,

For a example only ,i told u the last character, It can be any digit in
the whole number, So i want to compare the whole 20 digit number with
other 20 digit number.

I want know ,why is it not comparing correctly. Is there anyother way
to compare
 
B

balakrishnan.dinesh

HI Rob,

Thanks for your information.

Then there no way to compare this. My senario is like that,
what can i do to solve this problem.
Can u give me a suggestion.
 
R

Randy Webb

(e-mail address removed) said the following on 7/28/2006 2:51 AM:
Hi Randy,

For a example only ,i told u the last character, It can be any digit in
the whole number, So i want to compare the whole 20 digit number with
other 20 digit number.
Click to expand...

The value of a text input is a string. Do a string comparison and you
don't run into the 16 digit limit.
I want know ,why is it not comparing correctly. Is there anyother way
to compare
Click to expand...

Compare them as strings.

if (string1 == string2)
 
R

RobG

HI Rob,

Thanks for your information.

Then there no way to compare this. My senario is like that,
what can i do to solve this problem.
Can u give me a suggestion.
Click to expand...

An algorithm is:

- get the values as strings (input elements will return strings)
- check they are both integers
- remove leading zeros
- the longest one is bigger
- if they are the same length, check digit by digit
- if the numbers are really big, chec, blocks of say
10 digits at a time.

e.g. the following example returns true only if a and b are integers
and a is bigger than b. If a or b aren't integers, 'undefined' is
returned. Otherwise, if b is bigger or equal to a, false is returned.

function isBigger(a, b)
{
// Make sure they are integers
var re = new RegExp('\\D','g');
if (re.test(a) || re.test(b)){
return undefined;
}

// Trim leading zeros
a = a.replace(/^0+/g,'');
b = b.replace(/^0+/g,'');

// Check lengths
if (a.length != b.length){
return (a.length > b.length);
}

// Compare digits
var a = a.split('');
var b = b.split('');
for (var i=0, j=a.length; i<j; i++){
if (+a > b) return true;
}
return false;
}
 
R

Randy Webb

RobG said the following on 7/28/2006 5:07 AM:
An algorithm is:

- get the values as strings (input elements will return strings)
- check they are both integers
- remove leading zeros
Click to expand...

If they are defined in the HTML then you can be pretty assured whether
they are numbers or not, but even then, comparing them as strings is
simpler and easier. After removing the leading zeros, if they are equal
as strings they must be equal as numbers. If they aren't equal as
strings then they aren't equal as numbers.
 
P

Paul

RobG said:
HI Rob,

Thanks for your information.

Then there no way to compare this. My senario is like that,
what can i do to solve this problem.
Can u give me a suggestion.
Click to expand...

An algorithm is:

- get the values as strings (input elements will return strings)
- check they are both integers
- remove leading zeros
- the longest one is bigger
- if they are the same length, check digit by digit
- if the numbers are really big, chec, blocks of say
10 digits at a time.

e.g. the following example returns true only if a and b are integers
and a is bigger than b. If a or b aren't integers, 'undefined' is
returned. Otherwise, if b is bigger or equal to a, false is returned.

function isBigger(a, b)
{
// Make sure they are integers
var re = new RegExp('\\D','g');
if (re.test(a) || re.test(b)){
return undefined;
}

// Trim leading zeros
a = a.replace(/^0+/g,'');
b = b.replace(/^0+/g,'');

// Check lengths
if (a.length != b.length){
return (a.length > b.length);
}

// Compare digits
var a = a.split('');
var b = b.split('');
for (var i=0, j=a.length; i<j; i++){
if (+a > b) return true;
}
return false;
}
Click to expand...


Rob, I think you've got a great solution for the problem.
But, I'm going to offer up something a little more general purpose.
Inspired by Java's BigDecimal object. I'm going to propose a BigInteger
for java that will allow for comparison of really large numbers.
/*
* Representation of a big integer.
* the compare method returns a value greater than zero if the
comparing
* object is greater, a negative number if the object compared is
greater,
* or zero if the objects are equal.
*/
function BigInteger(strVal) {
this.parts = new Array();
for(var i=0; i<strVal.length; i++){
this.parts = +strVal.charAt(i);
}
this.compare = function(num){
var thisLength = this.parts.length;
var numLength = num.parts.length;
var greater = 0;
if(thisLength == numLength && thisLength>0){
var cnt = thisLength-1;
while( greater==0 && cnt >= 0 ){
greater = this.parts[cnt] - num.parts[cnt];
cnt--;
}
} else {
greater = (thisLength > numLength)?(1):(-1);
}
return greater;
}
}
Granted, creating an array of numbers is probably overkill for simple
comparison it does simplify extending to allow mathematical operations
pretty easily.

The original number check function could be modified to use this like:
function numberCheck()
{
var val1=document.getElementById('txt1').value;
var val2=document.getElementById('txt2').value;
var val1Bigint = new BigInteger(val1);
var val2Bigint = new BigInteger(val2);

var comp = val1Bigint.compare(val2Bigint);
if(comp==0) {
alert("Equal");
} else if(comp>0){
alert("Greater")
} else {
alert("Less");
}
}
 
D

Dr John Stockton

JRS: In article <[email protected]>
, dated Fri, 28 Jul 2006 02:07:30 remote, seen in
news:comp.lang.javascript said:
(e-mail address removed) wrote:
Click to expand...

That must seem very polite, to a Burmese.
An algorithm is:

- get the values as strings (input elements will return strings)
- check they are both integers
Click to expand...

But -79, +3, 123e45, 123e4567 are all integers. *Probably* he means
all-digit strings.
- remove leading zeros
Click to expand...

Probably. But 077 = 0x3F = 63 = 0.63e2, maybe.
function isBigger(a, b)
{
// Make sure they are integers
var re = new RegExp('\\D','g');
Click to expand...

I don't see that 'g' is needed, any non-digit fails an all-digit format.
if (re.test(a) || re.test(b)){
return undefined;
}

// Trim leading zeros
a = a.replace(/^0+/g,'');
b = b.replace(/^0+/g,'');
Ditto.

// Check lengths
if (a.length != b.length){
return (a.length > b.length);
}

// Compare digits
var a = a.split('');
var b = b.split('');
for (var i=0, j=a.length; i<j; i++){
if (+a > b) return true;
}
return false;
}
Click to expand...


It could be better to compare a.charAt(i) with b.charAt(i); it saves
generating length+length objects. Perhaps I mean charCodeAt throughout.
I suspect you've not handled b<a, though he seems not to need that.


I think I'd go lower-level.

Take strings A & B.

From the Subject line, we can take it that the numbers consist only of
decimal digits, to be interpreted as decimal (if doubtful, use a
preliminary RegExp [*]).

We can also take it that they are of equal length; and, if not, it's
easy to prepend a string of zeroes to the shorter. At this stage they
can in fact be compared as strings.

Otherwise, loop through both strings, subtracting A.charAt and
B.charAt. At the first difference, return that difference. Finally
return 0 (the final difference; so it'd be nice to use
Dif = cA-cB ; if (!Dif) break ; ... return Dif ; ).


ISTM that charAt should be swift given what ECMA seems to say about
all characters using 16 bits. If charAt[50] requires 100 or more bytes
to be considered, I'd not want to use it here.

ISTM that the strings could perhaps be broken up into ten-digit units
with a cunning RegExp (actually, cunning not needed where the length is
a multiple of 10) match, and the resulting strings compared numerically
in turn with unary +.
 

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