Webkit JS Core Bug in Math operation

https://bugs.webkit.org/show_bug.cgi?id=40367
When will Apple push out official Safari updates to fix this bug in Safari 4.1 and Safari 5.

HI,
Hard to say since the Apple Discussions TOU forbids discussions regarding when an update might be available.
Help and Terms of Use
Carolyn

Similar Messages

Math Operations on LV for Palm

I am trying to write a program for a Treo 680 running PalmOS 5.4 (Garnet) using LV for PDA 8.0, and I observe some strange behavior in math operations. The test that I did was to calculate B = log10(A). Both A and B are defined as DBL. The B field is programmed to show 16 decimal places. When I run this program on the PC, I get the following results:
log10(10) = 1.0000000000000000
log10(50) = 1.6989700043360187
log10(100)= 2.0000000000000000
When I run the same program on the Treo (or on Palm OS Garnet Simulator - same results), I get
log10(10) = 1.0000000000
log10(50) = 1.2694732747
log10(100)= 2.0000000000
A couple more interesting results:
log10(72) = 1.4278357668
log10(73) = 1.0043294009
Two questions:
1) Why do the results on Palm have only 10 decimal places?
2) Why are non-integer results so far from the correct values?
A few notes:
1) This is not a question of DBL vs. SGL: log10(50) in SGL arithmetic is 1.6989699602127075
on the PC - the difference starts only in the 7th decimal place.
2) I have MathLib.prc installed on the Treo (and on the simulator).
3) The Palm calculator works correctly: e.g. log10(50) = 1.698970004336
TIA,
Sergey

Cepera wrote:
I mean, how can you have a bug in such a fundamental operation as log(a)?
As mentioned, the PDA module converts your code to C, which is an error prone process.
Couple that with the fact that the PDA module has a relatively small market (and the Palm version an even smaller one, as evidenced by its scrubbing), and you can see how bugs like this can slip through the cracks. As you mentioned, the error does not occur for all values of a.
Try to take over the world!

Math operations on history chart

Hi everyone, I'd like to be able to do some math processing to chart history in that way:
1) visualizing continuous data from serial port on a multi plot chart ---> done
2) at a certain point be able to do some basic math operation on all the chart plots, say for example plot1 multiply by 2, plot 2 multiply by 1.5 ecc.. and visualize those operations on all the history collected till that moment on the same chart replacing raw data, and maybe return to visualize all the raw data collected, programmatically.
Thanx

Hi ghrsdr,
The best way to do it is to maintain a data structure in memory containing a portion of the acquired data (even all). When you need to do some operations on the data you could use this data structure to retrieve and modify your data, even creating a new data structure to maintain the original data. If the data set is large you may consider to stream data to and from your hard disk. The chart should be used only for data visualization purposes, not as a data structure to store the acquired data.
I suggest you to read the following white papers on the management of large data sets in LabVIEW.
Managing Large Data Sets in LabVIEW
http://www.ni.com/white-paper/3625/en
Memory Management for Large Data Sets
http://zone.ni.com/reference/en-XX/help/371361J-01/lvconcepts/memory_management_for_large_data_sets/
I hope I have given you some useful suggestion for your project.
Best regards,
Enrico

Deprecated math operator ^ ?

I was using Cold Fusion MX7, and now I'm on CF8.
This week someone told me that I had a problem with one of my
calculation, it use to be giving the good result but now it is not
giving good result. The only thing I can see is that the math
operator ^ is deprecated in CF8, is there something that replace
it???

> the math operator ^ is deprecated in CF8,
Cite?
It certainly doesn't say anything like that in the docs:
http://livedocs.adobe.com/coldfusion/8/htmldocs/Expressions_03.html
{quote}
^
Exponentiation: Return the result of a number raised to a
power (exponent).
Use the caret character (^) to separate the number from the
power; for
example, 2^3 is 8. Real and negative numbers are allowed for
both the base
and the exponent. However, any expression that equates to an
imaginary
number, such -1^.5 results in the string "-1.#IND. ColdFusion
does not
support imaginary or complex numbers.
{quote}
But anyway, if something is deprecated, it still works same
as it ever did,
it's simply flagged for obsolescence. So even if the ^
operator was
deprecated, there'd be no difference in how it worked before
and after
deprecation.
CF seldom deprecates things, and even when deprecated, things
are seldom
actually obsoleted. A good example is the parameterExists()
function: it's
be deprecated since (at least) CF5. It's still there and
functional in CF8
though.
> <cfset mtp = #rem# * 100 * ( (1 - ((1+
(#tme#/100))^-#duree#) ) ) / #tme#>
Well you don't need those pound signs in there, but having
them just makes
your code look untidy, it doesn't actually impact anything.
What kind of differences in results are you seeing?
Adam

I went to the apple store because i had a bug on the operating system on my iphone 5, they updated the operating system but they told me not to restore my phone, can I restore my pictures and contacts or will it bring the bug back again?

i was having problems with my phone so i went into the apple store and it turns out there was a problem with my operating systerm so i got an update in the store. I want to get my contacts and my pictures back from my old iphone but when I was in the apple shop the man told me to start the phone as a new phone instead of restoring it from an earlier back up as i could get the bug back once restored, is there any way for me to get my pictures and contacts back without restoring it and getting the bug? Thanks

You should restore from the back up so you can get the info back on your device. Then, what you can do is import those pictures to your computer.
At that point, you'll want to restore again as new possibly, then once your phone is as new, use iTunes to sync those picture from your computer back to your phone.
As far as contacts, do you know if those were ever backed up to iCloud? If so, all you should need to do is sign in with you iCloud account.

Bug: Context menu operations on multiple selection of uix-nodes do not work

Hi,
when i select multiple .uix files in the navigator and then chose "Check XML Syntax", "Validate" or "Auto Indent XML" from the context menu, the selected operation is only done for the first file in the selection.
This is a nice Bug because this works for Make and Rebuild context menuitems, even if those items display the filename even for multiple selection ("Make AFile.uix).
Would be nice if those operations would work on the "HTML Sources" node or a subfolder.
Regards, Markus

thanks. we'll try to fix this in an upcoming release.

(Error in documentation?) Math operator precedence problem

Hello,
I apologize in advance, while I do know programming (I'm a PHP and Perl programmer) I'm fairly new to Java so this may well be a silly question, but that's why I am here. I hope you'll show me where my error is so I can finally set this to rest before it drives me mad :)
(Also, I hope I'm posting this in the right forum?)
So, I am taking a Java class, and the question in the homework related to operand precendence. When dealing with multiplication, division and addition, things were fine, the documentation is clear and it also makes sense math-wise (multiplication comes before addition, etc etc).
However, we got this exercise to solve:
If u=2, v=3, w=5, x=7 and y=11, find the value assuming int variables:
u++ / v+u++ *w
Now, according to the operator precedence table (http://docs.oracle.com/javase/tutorial/java/nutsandbolts/operators.html) the unary operator u++ comes first, before multiplication, division and addition.
This would mean I could rewrite the exercise as:
((u+1)/v) + ((u+1)*w) = (3/3) + (4*5) = 1+20 = 21
However, if I run this in Java, the result I get is 15.
I tried breaking up the result for the two values in the Java code, so I could see where the problem is with my calculation.
For
System.out.println(u++ /v);
I get 0
For
System.out.println("u++ *w");
I get 15
My professor suggested I attempt to change the values from int to float, so I can see if the division came out to be something illogical. I did so, and now I get:
For
System.out.println(u++ /v);
I get 0.66667
For
System.out.println("u++ *w");
I get 15.0000
Which means that for the first operation (the division) the division happens on 2/3 (which is 0.6667) and only afterwards the u value is increased. That is, the u++ operation is done after the division, not before. The same happens with the multiplication; when that is carried out, the value of u is now 3 (after it was raised by one in the previous operation) and the first to be carried out is the multiplication (3*5=15) and only then the u++.
That is entirely against the documentation.
This is the script I wrote to test the issue:
public class MathTest {
     public static void main (String [] args) {
          float u=2, v=3, w=5, x=7, y=11;
          System.out.println("Initial value for 'u': " + u);
          float tmp1 = u++ /v;
          System.out.println("u++ /v = " + tmp1);
          System.out.println("First ++ value for 'u': " + u);
          float tmp2 = u++ *w;
          System.out.println("u++ *w= " + tmp2);
          System.out.println("Second ++ value for 'u': " + u);
          System.out.println(tmp1+tmp2);
The output:
Initial value for 'u': 2.0
u++ /v = 1.0
First ++ value for 'u': 3.0
u++ *w= 20.0
Second ++ value for 'u': 4.0
21.0
Clearly, the ++ operation is done after the division and after the multiplication.
Am I missing something here? Is the documentation wrong, or have I missed something obvious? This is driving me crazy!
Thanks in advance,
Mori

>
The fact that u++ is evaluated but in the calculation itself the "previously stored" value of u is used is completely confusing.
>
Yes it can be confusing - and no one is claiming otherwise. Here are some more thread links from other users about the same issue.
Re: Code Behavior is different in C and Java.
Re: diffcult to understand expression computaion having operator such as i++
Operator Precedence for Increment Operator
>
So, when they explain that ++ has a higher precedence, the natural thing to consider is that you "do that first" and then do the rest.
>
And, as you have discovered, that would be wrong and, to a large degree, is the nut of the issue.
What you just said illustrates the difference between 'precedence' and 'evaluation order'. Precedence determines which operators are evaluated first. Evaluation order determines the order that the 'operands' of those operators are evaluated. Those are two different, but related, things.
See Chapter 15 Expressions
>
This chapter specifies the meanings of expressions and the rules for their evaluation.
>
Sections in the chapter specify the requirements - note the word 'guarantees' in the quoted text - see 15.7.1
Also read carefully section 15.7.2
>
15.7. Evaluation Order
The Java programming language guarantees that the operands of operators appear to be evaluated in a specific evaluation order, namely, from left to right.
15.7.1. Evaluate Left-Hand Operand First
The left-hand operand of a binary operator appears to be fully evaluated before any part of the right-hand operand is evaluated.
If the operator is a compound-assignment operator (§15.26.2), then evaluation of
the left-hand operand includes both remembering the variable that the left-hand
operand denotes and fetching and saving that variable's value for use in the implied
binary operation.
15.7.2. Evaluate Operands before Operation
The Java programming language guarantees that every operand of an operator (except the conditional operators &&, ||, and ? appears to be fully evaluated before any part of the operation itself is performed.
15.7.3. Evaluation Respects Parentheses and Precedence
The Java programming language respects the order of evaluation indicated explicitly by parentheses and implicitly by operator precedence.
>
So when there are multiple operators in an expression 'precedence' determines which is evaluated first. And, the chart in your link shows 'postfix' is at the top of the list.
But, as the quote from the java language spec shows, the result of 'postfix' is the ORIGINAL value before incrementation. If it makes it easier for then consider that this 'original' value is saved on the stack or a temp variable for use in the calculation for that operand.
Then the evalution order determines how the calculation proceeds.
It may help to look at a different, simpler, but still confusing example. What should the value of 'test' be in this example?
i = 0;
test = i++; The value of 'test' will be zero. The value of i++ is the 'original' value before incrementing and that 'original' value is assigned to 'test'.
At the risk of further confusion here are the actual JVM instructions executed for your example. I just used
javap -c -l Test1to disassemble this. I manually added the actual source code so you can try to follow this
        int u = 2;
   4:iconst_2
   5:istore 5
        int v = 3;
   7:iconst_3
   8:istore          6
        int w = 5;
10:iconst_5
11:istore          7
        int z;
        z = z = u++ / v + u++ * w;
   13: iload   5
   15: iinc    5, 1
   18: iload   6
   20: idiv
   21: iload   5
   23: iinc    5, 1
   26: iload   7
   28: imul
   29: iadd
   30: dup
   31: istore 8
   33: istore 8The Java Virtual Machine Specification has all of the JVM instructions and what they do.
http://docs.oracle.com/javase/specs/jvms/se7/jvms7.pdf
Your assignment to z and formula starts with line 13: above
13: iload 5 - 'Load int from local variable' page 466
15: iinc 5, 1 - 'Increment local variable by constant' page 465
18: iload 6 - another load
20: idiv - do the division page
21: iload 5 - load variable 5 again
23: iinc 5, 1 - inc variable 5 again
26: iload 7 - load variable 7
28: imul - do the multiply
29: iadd - do the addition
30: dup - duplicate the top stack value and put it on the stack
31: istore 8 - store the duplicated top stack value
33: istore 8 - store the original top stack value
Note the description of 'iload'
>
The value of the local variable at index
is pushed onto the operand stack.
>
IMPORTANT - now note the description of 'iinc'
>
The value const is first sign-extended to an int, and then
the local variable at index is incremented by that amount.
>
The 'iload' loads the ORIGINAL value of the variable and saves it on the stack.
Then the 'iinc' increments the VARIABLE value - it DOES NOT increment the ORIGINAL value which is now on the stack.
Now note the description of 'idiv'
>
The values are popped
from the operand stack. The int result is the value of the Java
programming language expression value1 / value2. The result is
pushed onto the operand stack.
>
The two values on the stack include the ORIGINAL value of the variable - not the incremented value.
The above three instructions explain the result you are seeing. For postfix the value is loaded onto the stack and then the variable itself (not the loaded original value) is incremented.
Then you can see what this line does
21: iload   5 - load variable 5 againIt loads the variable again. This IS the incremented value. It was incremented by the 'iinc' on line 15 that I discussed above.
Sorry to get so detailed but this question seems to come up a lot and maybe now you have enough information and doc links to explore this to any level of detail that you want.

Is this a bug in table operator

I have an Oracle function that is invoked by the select statement as follow:
select count(*)
into i
from table(
select depts -- depts is a varray of dept_obj
from table(company_package.f_getCompany(k))
the function f_getCompany(k) gets invoked twice.
but if I change the SQL to
select depts
into dept_array
from table(company_package.f_getCompany(k));
then function f_getCompany(k) will be invoked just once.
Is this a bug within Oracle ? How can I get around that, I just want the function execute ONCE.
Any help would be appricated.
Thanks
Here are the rest of the code
create or replace type company_obj as object
company_id number,
company_name varchar2(20),
depts dept_arr
CREATE OR REPLACE
type company_arr is VARRAY(10000) of company_obj;
create or replace type dept_obj as object
dept_id number,
dept_name varchar2(30)
CREATE OR REPLACE
type dept_arr is VARRAY(10000) of dept_obj;
create or replace package company_package is
number_of_times int := 0;
function f_getCompany(company_id number) return company_arr;
end company_package;
create or replace package body company_package is
function f_getCompany(company_id number) return company_arr is
comp_array company_arr;
comp_object company_obj;
dept_array dept_arr;
dept_object dept_obj;
begin
dept_object := dept_obj(100, 'Dept 1');
dept_array := dept_arr();
dept_array.extend;
dept_array(1) := dept_object;
dept_object := dept_obj(110, 'Dept 2');
dept_array.extend;
dept_array(2) := dept_object;
comp_object := company_obj(1, 'Company A', dept_array);
comp_array := company_arr();
comp_array.extend;
comp_array(1) := comp_object;
number_of_times := number_of_times + 1;
dbms_output.put_line('number of times = ' || number_of_times);
return comp_array;
end;
end company_package;
----------------------------------------------------------------------------------------------------

I noticed that myself in our project.
Our varchars2 are defined as VARCHAR2(xxx CHAR) - OWB puts the size*4
In fact if you have special characters like umlauts (ü,ä,ö,...) it will use 4 bytes per character.
You can try it yourself. Define a Varchar2(1 CHAR) and manually change the size of the Column in your mapping inside OWB (in filters, joins or your target table).
Then shoot an umlaut through the mapping and will end up with a "too small" error.
Dont mind the size*4 issue - we totally ignored it and run without error since 4 years now.

Simple math operation - Help needed

Hi, i have a simple mathematical operation to do, related to
a shooping cart, which i want to keep simple.
I have a input text box named QT1, where customers indicate
quantity
I have a dymanic text box, named ST1, where i want the value
of QT1 to be multiplied by 10$ (hence 10)
on the release of the button. I have a NaN answer. Here is
the code.
on (release) {
var qt1:Number;
st1 = qt1 * 10;
And then, a grand total button will add ST1 + ST2 + ST3, with
GT the name of the grand total dymamic box:
on (release) {
gt = st1 + st2 + st3;
The second part works, but the first one, with QT1 is giving
me a Nan (not a number) answer.
Any help appreciated.

Problem solved. That cary Auto-kern thing....

Math operator for pl/sql

Does anyone know wich operator may I use for calculating the module like in the following expression?
SELECT * FROM sale WHERE Customer_Id <operator> 2 = 0
Thanks!

Assuming you mean the modulo,
WHERE mod(customer_id,2) = 0Justin

Math operations on array in TestStand

Hi.
I have 2 questions
1. Is it possible to multiply an array by a number in TestStand?
I know there are some numeric functions available in TestStand but I don't know if like LabVIEW I can multiple ( subtract/add/divide...) all the elements of an array by a number
2. Let's say I want to pass a 1D array of numbers ( size 10) to a code module but I would like to just pass the first 5 elements . How can I do it in TestStand ? I have checked the array function but I couldn't find something like Array Subset?
Could you please help me with these question

In TS 2012 and later they added some new array functions such as Min, Max, Split, Sort, etc... They also added the ability to do Locals.MyArray[4..8]. Which will give you elements 4 through 8 of the 0 indexed array.
As for the adding to each number I don't think that capability exists. If it did it would be something like Locals.MyArray + 100. It defnitely doesn't work in TS 2010.
Hope this helps,
jigg
CTA, CLA
teststandhelp.com
~Will work for kudos and/or BBQ~

Math Addition Operator Question

Hi there,
I'm trying to write a very simple math equation:
var gal_width = 100;
var m_box_pos = 2000;
var total = (gal_width+m_box_pos);
When I display this in a text element, the numbers are shown together rather than being added together
(eg: 1002000)
I get the feeling Edge is interpreting the + as a string operator rather than a math operator.
Replacing the + with a - (as a test), the 2 numbers are being subtracted.
Multiplying the numbers with the * operator also works as it should.
I've tried many configerations, escaping the operator with no luck.
Any ideas of how to get addition to work?
Thanks

Hi 29sketc.
console.log( sym.$("text_111").text());
// logs : 111
console.log(typeof sym.$("text_111").text());
// logs : string
var number_222 = 222;
console.log(typeof number_222);
// logs : number
console.log( sym.$("text_111").text() +number_222);
// logs : 111222
text() has produced a String so + is interpreted here as string concatenation.
var number_111 = 111;
console.log(typeof number_111);
// logs : number
console.log(number_111 +number_222);
// logs : 333
When you have two Number variables + is interpreted as addition.
Gil

Now operating system for iPad won't let me post photos in Facebook. The sent and upgrade to fix bugs but it freezes in my settings and won't launch

bug in new operating system for iPad
fix stalls and freezes in Settings

You are not speaking to Apple here. We are all users like yourself. we cannot "fix" anything. Please describe which Settings area you are having problems with, what exactly happens and what, if any, error messages you receive.
one thing that cures ,any system errors is to reset the iPad. Press and hold both the Home button and the Sleep/Wake button continuously until the Apple logo appears. Then release the button and let the device restart. You will not lose data doing this. It's like a computer reboot.

Ical bug: updates to outlook events with iCloud sync do not work

I have a bug that needs to be submitted to Apple Developers. I am not sure that this is the correct forum, but here goes.
I have a calendar that syncs from my Mac to iCloud. If I get a calendar invite from someone using outlook, everything works fine. I accept the invite and it is added to ical (6) and syncs with the iCloud calendar.
Now if I get an update to the invite - especially a change of time, the local version on my mac updates to the changed event. Up to now everything is fine, but as soon as iCal syncs with iCloud, the iCloud server thinks that I am trying to change an event that someone else is the organiser of. It then disallows the update. This in turn pops an error up on my mac - 'The server responded: “403” to operation CalDAVWriteEntityQueueableOperation'.
I have looked for workarounds/fixes but believe that none exist and that this is a core bug with iCal/iCloud sync and the way it tries to prevent updates to meetings organised by another party.

Hello vmepani
maybe I can suggest to you a way to circumvent the problem.
Why are you keeping a local Calendar copy under "my Mac"?
I do have my calendar shared through iCloud between my Mac Mountain Lion, the iPhone and the iPad (both alread IOS7) but don't need to keep any local copy of it in neither device. Anyway I'm able to see the calendar on each device even if I'm offline.
That's way I never had the problem you mentioned.
Not sure if this would work for you as well.
cheers
gg

Outlook 2010 / Exchange 2013: The operation failed. An object could not be found.

Both KB2956128 and KB2956203 causing issue. Removing the patch fixes the issue. I need to know what the best way to remove the patch remotely is. Maybe a SCCM job with this:
msiexec /I {90140000-0012-0000-0000-0000000FF1CE}MSIPATCHREMOVE={DEFF916D-4268-49CF-8FF3-E26253582E13}
msiexec /i {90140000-001A-0409-0000-0000000FF1CE}MSIPATCHREMOVE={DEFF916D-4268-49CF-8FF3-E26253582E13}
Please advise. Also looking for time frame for a fix to be published or these updates are rolled back/pulled by Microsoft. Thank you.
Also this is being reported by multiple people with no resolution. Example:
http://community.spiceworks.com/topic/790986-outlook-2010-bug-the-attempted-operation-failed-an-object-could-not?page=1

Hi,
Thank you for your feedback and apologize for the inconvenience caused.
We have heard several reports regarding issues caused by KB2956128. We've escalated this issue, but we're not sure whether and when there will be an hotfix for this issue. Please look out for our official blog articles and install
all available updates released.
Thanks for your understanding and support.
Steve Fan
TechNet Community Support
It's recommended to download and install
Configuration Analyzer Tool (OffCAT), which is developed by Microsoft Support teams. Once the tool is installed, you can run it at any time to scan for hundreds of known issues in Office
programs.
Please remember to mark the replies as answers if they help, and unmark the answers if they provide no help. If you have feedback for TechNet Support, contact
[email protected]

Webkit JS Core Bug in Math operation

Similar Messages

Maybe you are looking for