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  • How to change time (Advanced Eastern Time) on Slackware 8.1

    - by r0ca
    Hi all, I have a linux (Slackware) machine and the time/date is like, June 23rd 2003, 10:00am (It's 11 here) and I am not able to set the time to have it correct. I change the timezome to Montreal but the time is still wrong. Is there a way to force it to sync with my domain controler or even another online NTP server? Thanks, David.

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  • C# select elements from IList

    - by user313884
    I have a list of objects: IList<O> O has several properties but only two of them are relevant: Date and Duration I want to "split" the list into several lists that contain only the objects that have matching Date and Duration Properties. Example: InitialList: 0- Date==1, Duration==7 1- Date==1, Duration==7 2- Date==2, Duration==7 3- Date==2, Duration==7 4- Date==2, Duration==14 5- Date==2, Duration==14 Desired result: IList<IList<O>: 0- 0- Date==1, Duration==7 1- Date==1, Duration==7 1- 0- Date==2, Duration==7 1- Date==2, Duration==7 2- 0- Date==2, Duration==14 1- Date==2, Duration==14 i know this can be done with some linq selects but am not sure how. thanks for the help.

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  • Which platform can we expect one's complement being used there?

    - by Jian Lin
    For some questions such as checking whether a number is odd or even, I noted the comment, a & 1 won't work when it is a one's complement machine or when the code is ported to a platform that uses one's complement. Since 30 years ago on the Superboard, TRS-80, Apple II, I haven't seen a system with one's complement. Are there popular systems that use one's complement still, or do we have some cell phone or mobile device that uses one's complement?

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  • Why does the minus operator give different result than the TIMESTAMPDIFF() function in mysql?

    - by f3r3nc
    Since TIMESTAMP in mysql is stored as a 32bit value representing the time interval from 1970-jan-1 0:00:00 in seconds, I assumed that using minus (-) operator on TIMESTAMP values would give the difference of these values in seconds. Actually not: +---------------------------------------------------------------------+ | TIMESTAMP("2010-04-02 10:30:00") - TIMESTAMP("2010-04-02 10:29:59") | +---------------------------------------------------------------------+ | 41.000000 | +---------------------------------------------------------------------+ 1 row in set (0.05 sec) mysql> select timestampdiff(SECOND,TIMESTAMP("2010-04-02 10:30:00"),TIMESTAMP("2010-04-02 10:29:59")); +-----------------------------------------------------------------------------------------+ | timestampdiff(SECOND,TIMESTAMP("2010-04-02 10:30:00"),TIMESTAMP("2010-04-02 10:29:59")) | +-----------------------------------------------------------------------------------------+ | -1 | +-----------------------------------------------------------------------------------------+ mysql> select TIMESTAMP("2010-04-02 10:30:00") - TIMESTAMP("2010-04-02 10:30:01") ; +---------------------------------------------------------------------+ | TIMESTAMP("2010-04-02 10:30:00") - TIMESTAMP("2010-04-02 10:30:01") | +---------------------------------------------------------------------+ | -1.000000 | +---------------------------------------------------------------------+ +---------------------------------------------------------------------+ | TIMESTAMP("2010-04-02 10:30:00") - TIMESTAMP("2010-04-02 10:31:00") | +---------------------------------------------------------------------+ | -100.000000 | +---------------------------------------------------------------------+ It seems like one minute difference is 100 instead of 60. Why is this?

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  • Shift count negative or too big error - correct solution?

    - by PeterK
    I have the following function for reading a big-endian quadword (in a abstract base file I/O class): unsigned long long CGenFile::readBEq(){ unsigned long long qT = 0; qT |= readb() << 56; qT |= readb() << 48; qT |= readb() << 40; qT |= readb() << 32; qT |= readb() << 24; qT |= readb() << 16; qT |= readb() << 8; qT |= readb() << 0; return qT; } The readb() functions reads a BYTE. Here are the typedefs used: typedef unsigned char BYTE; typedef unsigned short WORD; typedef unsigned long DWORD; The thing is that i get 4 compiler warnings on the first four lines with the shift operation: warning C4293: '<<' : shift count negative or too big, undefined behavior I understand why this warning occurs, but i can't seem to figure out how to get rid of it correctly. I could do something like: qT |= (unsigned long long)readb() << 56; This removes the warning, but isn't there any other problem, will the BYTE be correctly extended all the time? Maybe i'm just thinking about it too much and the solution is that simple. Can you guys help me out here? Thanks.

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  • Custom model in ASP.NET MVC controller: Custom display message for Date DataType

    - by Rita
    Hi I have an ASP.NET MVC Page that i have to display the fields in customized Text. For that I have built a CustomModel RequestViewModel with the following fields. Description, Event, UsageDate Corresponding to these my custom Model has the below code. So that, the DisplayName is displayed on the ASP.NET MVC View page. Now being the Description and Event string Datatype, both these fields are displaying Custom DisplayMessage. But I have problem with Date Datatype. Instead of "Date of Use of Slides", it is still displaying UsageDate from the actualModel. Anybody faced this issue with DateDatatype? Appreciate your responses. Custom Model: [Required(ErrorMessage="Please provide a description")] [DisplayName("Detail Description")] [StringLength(250, ErrorMessage = "Description cannot exceed 250 chars")] // also need min length 30 public string Description { get; set; } [Required(ErrorMessage="Please specify the name or location")] [DisplayName("Name/Location of the Event")] [StringLength(250, ErrorMessage = "Name/Location cannot exceed 250 chars")] public string Event { get; set; } [Required(ErrorMessage="Please specify a date", ErrorMessageResourceType = typeof(DateTime))] [DisplayName("Date of Use of Slides")] [DataType(DataType.Date)] public string UsageDate { get; set; } ViewCode: <p> <%= Html.LabelFor(model => model.Description) %> <%= Html.TextBoxFor(model => model.Description) %> <%= Html.ValidationMessageFor(model => model.Description) %> </p> <p> <%= Html.LabelFor(model => model.Event) %> <%= Html.TextBoxFor(model => model.Event) %> <%= Html.ValidationMessageFor(model => model.Event) %> </p> <p> <%= Html.LabelFor(model => model.UsageDate) %> <%= Html.TextBoxFor(model => model.UsageDate) %> <%= Html.ValidationMessageFor(model => model.UsageDate) %> </p>

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  • DateTime: Require the user to enter a time component

    - by Heinzi
    Checking if a user input is a valid date or a valid "date + time" is easy: .NET provides DateTime.TryParse (and, in addition, VB.NET provides IsDate). Now, I want to check if the user entered a date including a time component. So, when using a German locale, 31.12.2010 00:00 should be OK, but 31.12.2010 shouldn't. I know I could use DateTime.TryParseExact like this: Dim formats() As String = {"d.M.yyyy H:mm:ss", "dd.M.yyyy H:mm:ss", _ "d.MM.yyyy H:mm:ss", "d.MM.yyyy H:mm:ss", _ "d.M.yyyy H:mm", ...} Dim result = DateTime.TryParseExact(userInput, formats, _ Globalization.CultureInfo.CurrentCulture, ..., result) but then I would hard-code the German format of specifying dates (day dot month dot year), which is considered bad practice and will make trouble should we ever want to localize our application. In addition, formats would be quite a large list of all possible combinations (one digit, two digits, ...). Is there a more elegant solution?

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  • simple c# arythmetics. winForms

    - by jello
    I'm doing simple divisions in c#, and I am a bit puzzled by its intricacies. Here's some code, and in the comments, the result. (btw, I only compile with 1 line not commented, if you say that I have 5 declarations of the same variable) double result = 2 / 3; //gives 0 double result = Convert.ToDouble(2) / Convert.ToDouble(3); // is good double result = double.Parse(2) / double.Parse(3); // gives me errors double result = double.Parse(2 / 3); // gives me errors double result = Convert.ToDouble(2 / 3); // gives 0 MessageBox.Show(result.ToString()); so if you have a bunch of integers you wanna mess with, you have to convert each one to a double. pretty tedious...

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  • simple c# arithmetics. winForms

    - by jello
    I'm doing simple divisions in c#, and I am a bit puzzled by its intricacies. Here's some code, and in the comments, the result. (btw, I only compile with 1 line not commented, if you say that I have 5 declarations of the same variable) double result = 2 / 3; //gives 0 double result = Convert.ToDouble(2) / Convert.ToDouble(3); // is good double result = double.Parse(2) / double.Parse(3); // gives me errors double result = double.Parse(2 / 3); // gives me errors double result = Convert.ToDouble(2 / 3); // gives 0 MessageBox.Show(result.ToString()); so if you have a bunch of integers you wanna mess with, you have to convert each one to a double. pretty tedious...

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  • How to make strtotime parse dates in Australian (i.e. UK) format: dd/mm/yyyy?

    - by Iain Fraser
    I can't beleive I've never come across this one before. Basically, I'm parsing the text in human-created text documents and one of the fields I need to parse is a date and time. Because I'm in Australia, dates are formatted like dd/mm/yyyy but strtotime only wants to parse it as a US formatted date. Also, exploding by / isn't going to work because, as I mentioned, these documents are hand-typed and some of them take the form of d M yy. I've tried multiple combinations of setlocale but no matter what I try, the language is always set to US English. I'm fairly sure setlocale is the key here, but I don't seem to be able to strike upon the right code. Tried these: au au-en en_AU australia aus Anything else I can try? Thanks so much :) Iain Example: $mydatetime = strtotime("9/02/10 2.00PM"); echo date('j F Y H:i', $mydatetime); Produces 2 September 2010 14:00 I want it to produce: 9 February 2010 14:00

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  • Is there a way to do 'correct' arithmetical rounding in .NET? / C#

    - by Markus
    I'm trying to round a number to it's first decimal place and, considering the different MidpointRounding options, that seems to work well. A problem arises though when that number has sunsequent decimal places that would arithmetically affect the rounding. An example: With 0.1, 0.11..0.19 and 0.141..0.44 it works: Math.Round(0.1, 1) == 0.1 Math.Round(0.11, 1) == 0.1 Math.Round(0.14, 1) == 0.1 Math.Round(0.15, 1) == 0.2 Math.Round(0.141, 1) == 0.1 But with 0.141..0.149 it always returns 0.1, although 0.146..0.149 should round to 0.2: Math.Round(0.145, 1, MidpointRounding.AwayFromZero) == 0.1 Math.Round(0.146, 1, MidpointRounding.AwayFromZero) == 0.1 Math.Round(0.146, 1, MidpointRounding.ToEven) == 0.1 Math.Round(0.146M, 1, MidpointRounding.ToEven) == 0.1M Math.Round(0.146M, 1, MidpointRounding.AwayFromZero) == 0.1M I tried to come up with a function that addresses this problem, and it works well for this case, but of course it glamorously fails if you try to round i.e. 0.144449 to it's first decimal digit (which should be 0.2, but results 0.1.) (That doesn't work with Math.Round() either.) private double round(double value, int digit) { // basically the old "add 0.5, then truncate to integer" trick double fix = 0.5D/( Math.Pow(10D, digit+1) )*( value = 0 ? 1D : -1D ); double fixedValue = value + fix; // 'truncate to integer' - shift left, round, shift right return Math.Round(fixedValue * Math.Pow(10D, digit)) / Math.Pow(10D, digit); } I assume a solution would be to enumerate all digits, find the first value larger than 4 and then round up, or else round down. Problem 1: That seems idiotic, Problem 2: I have no idea how to enumerate the digits without a gazillion of multiplications and subtractios. Long story short: What is the best way to do that?

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  • How is schoolbook long division an O(n^2) algorithm?

    - by eSKay
    Premise: This Wikipedia page suggests that the computational complexity of Schoolbook long division is O(n^2). Deduction: Instead of taking "Two n-digit numbers", if I take one n-digit number and one m-digit number, then the complexity would be O(n*m). Contradiction: Suppose you divide 100000000 (n digits) by 1000 (m digits), you get 100000, which takes six steps to arrive at. Now, if you divide 100000000 (n digits) by 10000 (m digits), you get 10000 . Now this takes only five steps. Conclusion: So, it seems that the order of computation should be something like O(n/m). Question: Who is wrong, me or Wikipedia, and where?

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  • MYSQL query to get all entries with specific time, from PHP?

    - by meds
    I'm trying to query a mysql table which places its date in the following format: yyyy-mm-dd hh:mm:ss So it's date and time, and that's all in a single field. Now from php I want to get the time and query the table to only return entries where the date field is less than 24 hours old. I'm having issues with the system because PHPs get time seems to return the values seperately and I'm struggling to figure out how to make it work with mysql queries. This seems fairly simple but I'm quite new to php so sorry if I'm completely missing something..

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  • basic boost date_time input format question

    - by Chris H
    I've got a pointer to a string, (char *) as input. The date/time looks like this: Sat, 10 Apr 2010 19:30:00 I'm only interested in the date, not the time. I created an "input_facet" with the format I want: boost::date_time::date_input_facet inFmt("%a %d %b %Y"); but I'm not sure what to do with it. Ultimately I'd like to create a date object from the string. I'm pretty sure I'm on the right track with that input facet and format, but I have no idea how to use it. Thanks.

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  • Why subtract null pointer in offsetof()?

    - by Bruce Christensen
    Linux's stddef.h defines offsetof() as: #define offsetof(TYPE, MEMBER) ((size_t) &((TYPE *)0)->MEMBER) whereas the Wikipedia article on offsetof() (http://en.wikipedia.org/wiki/Offsetof) defines it as: #define offsetof(st, m) \ ((size_t) ( (char *)&((st *)(0))->m - (char *)0 )) Why subtract (char *)0 in the Wikipedia version? Is there any case where that would actually make a difference?

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  • ls output changing when used through exec()

    - by user359650
    I'm using the ls command via PHP and exec() and I get a different output than when I run the same command via the shell. When running ls through PHP the year and month of the date get changed into the month name: Running the command through the shell: $ ls -lh /path/to/file -rw-r--r-- 1 sysadmin sysadmin 36M 2011-05-18 13:25 file Running the command via PHP: <?php exec("ls -lh /path/to/file", $output); print_r($output); /* Array ( [0] => -rw-r--r-- 1 sysadmin sysadmin 36M May 18 13:25 file ) */ Please note that: -the issue doesn't occur when I run the PHP script via the cli (it only occurs when run through apache) -I checked the source code of the page to make sure that what I was seeing was what I was getting (and I do get the month name instead of the proper date) -I also run the ls command through the shell as the www-data user to see if ls was giving different output depending on the user (the output is the always the same from the shell, that is I get the date in yyyy-mm-dd instead of the month name)

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  • Picking good first estimates for Goldschmidt division

    - by Mads Elvheim
    I'm calculating fixedpoint reciprocals in Q22.10 with Goldschmidt division for use in my software rasterizer on ARM. This is done by just setting the nominator to 1, i.e the nominator becomes the scalar on the first iteration. To be honest, I'm kind of following the wikipedia algorithm blindly here. The article says that if the denominator is scaled in the half-open range (0.5, 1.0], a good first estimate can be based on the denominator alone: Let F be the estimated scalar and D be the denominator, then F = 2 - D. But when doing this, I lose a lot of precision. Say if I want to find the reciprocal of 512.00002f. In order to scale the number down, I lose 10 bits of precision in the fraction part, which is shifted out. So, my questions are: Is there a way to pick a better estimate which does not require normalization? Also, is it possible to pre-calculate the first estimates so the series converges faster? Right now, it converges after the 4th iteration on average. On ARM this is about ~50 cycles worst case, and that's not taking emulation of clz/bsr into account, nor memory lookups. Here is my testcase. Note: The software implementation of clz on line 13 is from my post here. You can replace it with an intrinsic if you want. #include <stdio.h> #include <stdint.h> const unsigned int BASE = 22ULL; static unsigned int divfp(unsigned int val, int* iter) { /* Nominator, denominator, estimate scalar and previous denominator */ unsigned long long N,D,F, DPREV; int bitpos; *iter = 1; D = val; /* Get the shift amount + is right-shift, - is left-shift. */ bitpos = 31 - clz(val) - BASE; /* Normalize into the half-range (0.5, 1.0] */ if(0 < bitpos) D >>= bitpos; else D <<= (-bitpos); /* (FNi / FDi) == (FN(i+1) / FD(i+1)) */ /* F = 2 - D */ F = (2ULL<<BASE) - D; /* N = F for the first iteration, because the nominator is simply 1. So don't waste a 64-bit UMULL on a multiply with 1 */ N = F; D = ((unsigned long long)D*F)>>BASE; while(1){ DPREV = D; F = (2<<(BASE)) - D; D = ((unsigned long long)D*F)>>BASE; /* Bail when we get the same value for two denominators in a row. This means that the error is too small to make any further progress. */ if(D == DPREV) break; N = ((unsigned long long)N*F)>>BASE; *iter = *iter + 1; } if(0 < bitpos) N >>= bitpos; else N <<= (-bitpos); return N; } int main(int argc, char* argv[]) { double fv, fa; int iter; unsigned int D, result; sscanf(argv[1], "%lf", &fv); D = fv*(double)(1<<BASE); result = divfp(D, &iter); fa = (double)result / (double)(1UL << BASE); printf("Value: %8.8lf 1/value: %8.8lf FP value: 0x%.8X\n", fv, fa, result); printf("iteration: %d\n",iter); return 0; }

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  • Pointer incrementing query

    - by Craig
    I have been looking at this piece of code, and it is not doing what I expect. I have 3 globals. int x, y, *pointer, z; Inside of main I declare them. x = 10; y = 25; pointer = &x; now at this point &x is 0x004A144 &y is 0x004A138 pointer is pointing to 0x004A144 Now when I increment: y = *++pointer; it points to 0x004A148, this is the address y should be at shouldn't it? The idea is that incrementing the pointer to 'x' should increment it to point at y, but it doesn't seem to want to declare them in in order like I expect. If this a VS2005 / 2008 problem? Or maybe an Express problem? This isn't really homework, as I have done it a couple of years ago but I was revising on my pointer stuff and I tried this again. But this time I am getting unexpected results. Does anyone have opinions on this? *UPDATE sorry should be more clear, 'thought' on declaration 'y' should be at 148, and that incrementing the pointer pointing to x should increment 'pointer' to 148 (which it does), but that isn't where y is. Why isn't y declaring where it should be.

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  • PHP - Large Interger mod calculation

    - by Kami
    I need to calculate modulus with large number like : <?php $largenum = 95635000009453274121700; echo $largenum % 97; ?> It's not working... beacause $largenum is too big for an int in PHP. Any idea how to do this ?

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  • php - comparing timestamp dates to make sure user is of minimum age

    - by Micheal Ken
    When a user signs up the system has to check that they are old enough to do so, in this example they have to be atleast 8 years old $minAge = strtotime(date("d")."-".date("m")."-".(date("Y")-8)); $dob = strtotime($day."-".$month."-".$year); $minAge = 01-03-2004, $dob = 01-02-2011 I basically need to make sure this person was born before 2004 but I want to know whether I have to convert the timestamps to do a comparison or whether there is a more efficient way. Any help is appreciated, thank you

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  • Scaffolding A model with an attribute of type datetime creates a 10 years range in the form

    - by b_ayan
    For a simple rails application ( 1.86 /2.3.5) , lets say I run a simple scaffold script/generate scaffold blog title:string content:text published:date When I open up the new / edit view for the blog controller in index/new.html.erb , I see that the drop down enabler for date select has a date range of 2005 - 2015 , i.e 5 years +/- I tried to change this default behavior by introducing this code f.date_select :entered, :start_year => 1970, :end_year => 2020 Apparently this has no impact to the behavior mentioned above. How do I increase the date_select range which seems to be default?

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