make a float from two integers

[Ike]

If I have two variables that contain integer values, and I want the second
one to actually be the decimal portion of a single, floating-point number),
e.g. 3 and 7 to be therefore 3.7, how can I do this, if I don;t know in
advance how many significant digits the 'second' number will have? That is,
I do not know if I will have, say, 3 and 7 for 3.7, or 8 and 48501 for
8.48501.

It seems there must be a way if I treat these as String values, but I don;t
really see how to do this. Thanks, Ike

[Lasse Reichstein Nielsen]

"Ike" <rx*@hotmail.com> writes:

If I have two variables that contain integer values, and I want the second
one to actually be the decimal portion of a single, floating-point number),
e.g. 3 and 7 to be therefore 3.7, how can I do this, if I don;t know in
advance how many significant digits the 'second' number will have?
You can't. How would you represent 4.06?
That is, I do not know if I will have, say, 3 and 7 for 3.7, or 8
and 48501 for 8.48501.
or 3 and 7 for 3.07 or 3.007.

You would be better off if you reversed the second number, i.e.,
32 and 45 became 32.54. Then you could have 100 and 200 becomming
100.002, and you could represent all numbers in the possible range.
It seems there must be a way if I treat these as String values, but I don;t
really see how to do this.

You could convert them to strings, but you would already have lost the
necessary information, because a number value cannot have prefixed zeros,
and you need this.
/L
--
Lasse Reichstein Nielsen - lr*@hotpop.com
DHTML Death Colors: <URL:http://www.infimum.dk/HTML/rasterTriangleDOM.html>
'Faith without judgement merely degrades the spirit divine.'

[Martin Honnen]

Ike wrote:

If I have two variables that contain integer values, and I want the second
one to actually be the decimal portion of a single, floating-point number),
e.g. 3 and 7 to be therefore 3.7, how can I do this, if I don;t know in
advance how many significant digits the 'second' number will have? That is,
I do not know if I will have, say, 3 and 7 for 3.7, or 8 and 48501 for
8.48501.

One possible way:

var n1 = 8;
var n2 = 48501;
var n3 = parseFloat(String(n1) + '.' + String(n2));

--

Martin Honnen
http://JavaScript.FAQTs.com/

[Dr John Stockton]

JRS: In article <44***********************@newsread4.arcor-online.net>,
dated Sat, 20 May 2006 15:45:55 remote, seen in
news:comp.lang.javascript, Martin Honnen <ma*******@yahoo.de> posted :

Ike wrote: Please give informative attributions, as preferred by USEFOR. You're
not posting in a German newsgroup here.

If I have two variables that contain integer values, and I want the second
one to actually be the decimal portion of a single, floating-point number), ^^^^^^^
You mean that it represents the fractional part. In Javascript, all
Numbers are IEEE Doubles, which are binary floats - see FAQ.
e.g. 3 and 7 to be therefore 3.7, how can I do this, if I don;t know in
advance how many significant digits the 'second' number will have? That is,
I do not know if I will have, say, 3 and 7 for 3.7, or 8 and 48501 for

You can do that; but you will be unable to represent a non-integer with
a zero following its decimal point.
One possible way:

var n1 = 8;
var n2 = 48501;
var n3 = parseFloat(String(n1) + '.' + String(n2));

or n3 = +(n1 + '.' + n2)
or n3 = n1 + n2/Math.pow(10, String(n2).length)

--
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[Thomas 'PointedEars' Lahn]

Martin Honnen wrote:

One possible way:

var n1 = 8;
var n2 = 48501;
var n3 = parseFloat(String(n1) + '.' + String(n2));

With the ',' operand, the other operands of the concatenation are
implicitly converted to string, so String() is not necessary.

var n3 = parseFloat(n1 + '.' + n2);
PointedEars
--
Those who desire to give up freedom in order to gain security,
will not have, nor do they deserve, either one.
-- Benjamin Franklin

[Dr John Stockton]

JRS: In article <14****************@PointedEars.de>, dated Thu, 25 May
2006 13:05:32 remote, seen in news:comp.lang.javascript, Thomas 'House-
Painter' Lahn <Po*********@web.de> posted :

Martin Honnen wrote:

One possible way:

var n1 = 8;
var n2 = 48501;
var n3 = parseFloat(String(n1) + '.' + String(n2));

With the ',' operand, the other operands of the concatenation are
implicitly converted to string, so String() is not necessary.

var n3 = parseFloat(n1 + '.' + n2);

You're posting too hastily, as usual, and not checking what you write.

There is no ',' operand.

Using parseFloat is unnecessary. See FAQ 4.21.

Your signature remains non-compliant with FYI28; fix it.
--
© John Stockton, Surrey, UK. ?@merlyn.demon.co.uk Turnpike v4.00 IE 4 ©
<URL:http://www.jibbering.com/faq/> JL/RC: FAQ of news:comp.lang.javascript
<URL:http://www.merlyn.demon.co.uk/js-index.htm> jscr maths, dates, sources.
<URL:http://www.merlyn.demon.co.uk/> TP/BP/Delphi/jscr/&c, FAQ items, links.