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From: Ry Nohryb on 31 May 2010 05:12
On May 30, 6:28 pm, Dr J R Stockton <reply1...(a)merlyn.demon.co.uk>
wrote:
> (...)
> Opera 10.10, but not Opera 10.53 : Number.toString(radix) ignores
> radix.

There's a bug in Safaris also:

(Math.PI/10000).toString(anyBase)
--> "0"
--
Jorge.
From: Evertjan. on 31 May 2010 05:32
Ry Nohryb wrote on 31 mei 2010 in comp.lang.javascript:

> String.prototype.toFP= function (base, n, r, w, div, s) {
> if (/[^0-9a-z\.+-]/i.test(this)) return NaN;
> n= this.split('.');
> if (isFinite(r= parseInt(n[0], base))) {
> if (n[1] && (w= n[1].length)) {
> /*trim until div is finite*/
> while (!isFinite(div= Math.pow(base, w))) w--;
> /*Tests sign properly for -0.xxxx*/
> s= (r || parseInt(n[0]+ "1", base)) < 0 ? -1:1;


s= parseInt(n[0]+ '1', base)<0 ? -1:1;


> r+= s* parseInt(n[1].substr(0, w), base)/ div;
> }
> }
> return r;
>};
>



--
Evertjan.
The Netherlands.
(Please change the x'es to dots in my emailaddress)
From: Evertjan. on 31 May 2010 05:49
Evertjan. wrote on 31 mei 2010 in comp.lang.javascript:

> Ry Nohryb wrote on 31 mei 2010 in comp.lang.javascript:
>
>> String.prototype.toFP= function (base, n, r, w, div, s) {
>> if (/[^0-9a-z\.+-]/i.test(this)) return NaN;
>> n= this.split('.');
>> if (isFinite(r= parseInt(n[0], base))) {
>> if (n[1] && (w= n[1].length)) {
>> /*trim until div is finite*/
>> while (!isFinite(div= Math.pow(base, w))) w--;
>> /*Tests sign properly for -0.xxxx*/
>> s= (r || parseInt(n[0]+ "1", base)) < 0 ? -1:1;
>
>
> s= parseInt(n[0]+ '1', base)<0 ? -1:1;

Or even:

s= n[0][0]=='-' ?-1 :1;

>> r+= s* parseInt(n[1].substr(0, w), base)/ div;
>> }
>> }
>> return r;
>>};
>>
>
>
>



--
Evertjan.
The Netherlands.
(Please change the x'es to dots in my emailaddress)
From: Stefan Weiss on 31 May 2010 07:05
On 31/05/10 09:15, Evertjan. wrote:
> Dr J R Stockton wrote on 30 mei 2010 in comp.lang.javascript:
>
>>>btw: in root 1 only the zero value can be expressed.
>
> radix for Latinists?
>
>> Eh?
>
> How would you express 5 or 0.7 ?

In a unary system it's more usual to count like this:

1
11
111
1111

Not all values can be expressed this way (for example, 0 or 1.05), but
at least you get more values that just the zero.


--
stefan
From: Evertjan. on 31 May 2010 07:29
Evertjan. wrote on 31 mei 2010 in comp.lang.javascript:

> Evertjan. wrote on 31 mei 2010 in comp.lang.javascript:
>
>> Ry Nohryb wrote on 31 mei 2010 in comp.lang.javascript:
>>
>>> String.prototype.toFP= function (base, n, r, w, div, s) {
>>> if (/[^0-9a-z\.+-]/i.test(this)) return NaN;


This test is not enough now, meseems:

a = '12345abc.5def'
document.write((a.toFP(10)) // 12345.0005

use:

var re = base>10 ?'a-'+String.fromCharCode(86+base) :'';
re = new RegExp('[^0-9'+re+'\.+-]','i');
if (re.test(this)) return NaN;

and add:

if (/[+-]/.test(this.substr(1))) return NaN; //+- must be in position 0
if (this.replace(/[^\.]+/g,'').length>1) return NaN; // count of . <=1



>>> n= this.split('.');
>>> if (isFinite(r= parseInt(n[0], base))) {
>>> if (n[1] && (w= n[1].length)) {
>>> /*trim until div is finite*/
>>> while (!isFinite(div= Math.pow(base, w))) w--;
>>> /*Tests sign properly for -0.xxxx*/
>>> s= (r || parseInt(n[0]+ "1", base)) < 0 ? -1:1;
>>
>>
>> s= parseInt(n[0]+ '1', base)<0 ? -1:1;
>
> Or even:
>
> s= n[0][0]=='-' ?-1 :1;
>
>>> r+= s* parseInt(n[1].substr(0, w), base)/ div;
>>> }
>>> }
>>> return r;
>>>};
>>>
>>
>>
>>
>
>
>



--
Evertjan.
The Netherlands.
(Please change the x'es to dots in my emailaddress)
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