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  • Server load spikes several times a day, load average for the past month is 5 times the load average all year

    - by AMF
    My Munin notifications set up for our (Debian) LAMP cluster have been notifying me continuously that our load on our production machine has been at dangerous levels. While the average load all year typically runs between 2 and 8, the load in the past month and only the past month -- has been skyrocketing to 10, 18, and occasionally even 50-60. The spikes last only 5-10 minutes at a time and occur about every 2-3 hours. The spikes do not effect performance only because I have a script that sends traffic off our server to a mirror CDN when the load goes above 10. I've looked for cron jobs that correlate with this timeframe but there is nothing I can see that would cause this. Site traffic is also normal (we receive about 200K visits per day). I'm also trying to think of anything I've changed around the time this problem began, and I really cannot think of anything. This is probably not much to go on. Maybe there is a clue in the top print-out (below) that I'm not seeing. How do I proceed to find the cause? -- Typical top when the load is NOT spiking: top - 11:13:09 up 472 days, 25 min, 1 user, load average: 6.08, 4.29, 3.80 Tasks: 105 total, 1 running, 104 sleeping, 0 stopped, 0 zombie Cpu(s): 41.2%us, 5.8%sy, 0.0%ni, 49.5%id, 2.7%wa, 0.1%hi, 0.7%si, 0.0%st Mem: 3369592k total, 2166980k used, 1202612k free, 559504k buffers Swap: 2650684k total, 1892k used, 2648792k free, 1129116k cached PID USER PR NI VIRT RES SHR S %CPU %MEM TIME+ COMMAND 32046 apache 15 0 36300 12m 9828 S 20 0.4 0:01.97 apache2 32679 apache 15 0 36568 13m 10m S 19 0.4 0:01.69 apache2 31441 apache 15 0 36616 13m 10m S 19 0.4 0:04.13 apache2 31477 apache 15 0 36596 13m 9.8m S 15 0.4 0:01.99 apache2 31993 apache 15 0 36876 16m 12m S 12 0.5 0:02.01 apache2 31782 apache 15 0 36836 14m 10m S 8 0.4 0:02.17 apache2 32198 apache 15 0 36536 13m 10m S 7 0.4 0:01.59 apache2 880 apache 15 0 36508 9708 6236 S 7 0.3 0:00.42 apache2 31945 apache 17 0 36876 16m 13m S 5 0.5 0:03.17 apache2 32197 apache 16 0 36636 10m 7504 S 5 0.3 0:02.70 apache2 32326 apache 15 0 37024 11m 7632 S 5 0.3 0:02.15 apache2 32565 apache 15 0 37280 13m 9.8m S 5 0.4 0:03.75 apache2 32676 apache 15 0 36896 16m 12m S 4 0.5 0:00.95 apache2 32678 apache 15 0 36536 12m 9692 S 4 0.4 0:02.27 apache2 974 apache 16 0 37064 9888 6016 D 4 0.3 0:00.13 apache2 32150 apache 16 0 36832 13m 10m S 3 0.4 0:01.74 apache2 31780 apache 16 0 36848 11m 7660 S 3 0.3 0:02.87 apache2

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  • JCP 2012 Award Nominations are now open!

    - by heathervc
    The 10th JCP Annual Awards Nominations are now open until 16 July 2012. Submit nominations to [email protected] or use form here. The Java Community Process (JCP) program celebrates success. Members of the community nominate worthy participants, Spec Leads, and Java Specification Requests (JSRs) in order to cheer on the hard work and creativity that produces ground-breaking results for the community and industry in the Java Standard Edition (SE), Java Enterprise Edition (EE), or Java Micro Edition (ME) platforms. The community gets together every year at the JavaOne conference to applaud in person the winners of three awards: JCP Member/Participant of the Year, Outstanding Spec Lead, and Most Significant JSR. This year’s unveiling will occur Tuesday evening, 2 October, at the Annual JCP Community Party held in San Francisco.  Nominate today...descriptions of the award categories for this year: JCP Member/Participant Of The Year - This award recognizes the corporate or individual member (either Member or Participant) who has made the most significant positive impact on the community in the past year. Leadership, investment in the community, and innovation are some of the qualities that EC Members look for in voting for this award. Outstanding Spec Lead - The role of Spec Lead is not an easy one, and the person who takes that responsibility must be, among other things, technically savvy, able to build consensus in spite of diverse corporate goals, and focused on efficiency and execution. This award recognizes the person who has brought together these qualities the best in the past year, in leading a JSR for the Java community (Java SE, Java EE or Java ME). Most Significant JSR - Specification development is key to the success of the JCP program and helps ensure we remain a fresh and vibrant community. This award recognizes the Spec Lead and Expert Group that have contributed (either in progress or final) the most significant JSR for the Java community (Java SE, Java EE or Java ME) in the past year.

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  • And the Winner Is ...

    - by Oracle OpenWorld Blog Team
    If you know excellent Oracle technologists, now's your chance to nominate them for an award. by Karen Shamban It’s possible to win an Oracle Excellence Award in one of 12 categories this year—nominations are open now through July 17, 2012. Winning customers and partners will be hosted at Oracle OpenWorld or JavaOne 2012, where they can meet with Oracle executives, network with peers, and be featured in an upcoming edition of an Oracle publication such as Oracle Magazine. This year’s Oracle Excellence Award categories are: •    CIO of the Year•    Database Administrator of the Year•    Eco-Enterprise Innovation•    Java Business Innovation•    Leadership•    Oracle Fusion Middleware Innovation•    Proactive Support Champion–Global•    Specialized Partner of the Year–Europe, Middle East, and Africa•    Specialized Partner of the Year–Global•    Specialized Partner of the Year–North America•    Technologist of the Year Learn more about each award and nominate a deserving candidate now! Go to the Oracle Excellence Awards information page for details.

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  • comparing rows on a mysql table

    - by user311324
    Ok here's the deal I got one table with a bunch of client information. Each client makes up to one purchase a year which is represented by an individual row. there's a column for the year and there's a column the contains a unique identifier for each client. What I need to do is to construct a query that takes last year and this year and shows me which clients were here made a purchase last year but not make a purchase this year. I also need to build a query that shows me which clients did not make a purchase last year and the year before last but did make a purchase this year.

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  • Are there code signing certificates cheaper than US$99 per year? [closed]

    - by gerryLowry
    K Software discounts Comodo code signing certificates to US$99 per year. In the past, I've seen Commodo code signing certificates for US$80. I'm excluding CAcert which AFAIK are FREE but are not covered by browsers like Internet Explorer AFAIK. QUESTION: What is the best price per year for a code signing certificate? Thank you ~~ gerry (lowry) Edit: **THIS SHOULD NOT HAVE BEEN CLOSED** from the FAQ: http://stackoverflow.com/faq ---------------------------- What kind of questions can I ask here? Programming questions, of course! As long as your question is: * detailed and specific <====== YES! * written clearly and simply <====== YES! * of interest to other programmers <====== YES! I've been programming for over 40 years. http://gerrylowryprogrammer.com/ I've taught computer programming at the community college level. I'm a Star level contributer to forums.asp.net. http://forums.asp.net/members/gerrylowry.aspx IMO, I've a very good idea what is of interest to other programmers. MORE INFORMATION http://en.wikipedia.org/wiki/Code_signing http://msdn.microsoft.com/en-us/library/ms537361%28VS.85%29.aspx also: via Google: code signing Ensuring the integrity of code and executables that one distributes is just as much about programming as is knowing how to flip bits in assembler, use delegates in C#, and use the BDD context/specification still of "test first testing".

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  • How to scrape Google SERP based on copyright year?

    - by Michael Mao
    Hi all: I know there must be ways to do this sort of things. I am not pro in RoR or Python, not even an expert in PHP. So my solution tends to be quite dumb: It uses a FireFox add-on called imarcos to scrape the target urls from Google SERP, and use PHP to store info into the database. At the very core of my workaround there lies a problem: How to specifically find target urls based on their copyright year? I mean, something like "copyright 1998-2006" in the footer is to be considered a target, but my search results are not 100% accurate. I used the following url to search : http://www.google.com.au/#hl=en&q=inurl:.com.au+intext:copyright+1995..2007+--2008+--2009&start=0&cad=b&fp=6a8119b094529f00 It reads : search for pages that have .com.au in URL and a copyright range from 1995 to 2007 exclude the year of 2008 or 2009. Starting position is 0, of course the offset can be changed. I've already done a dummy list and honestly I am not pleased with the result. That's mostly because I cannot find a way to restrict search terms in the exact order as they are entered into the search url. copyright can appear in anywhere on page and doesn't necessarily before the years, that's the current story. Is there a more clear way to sort out this? Oh, almost forgot to say the client doesn't wanna spent too much in this - I cannot persuade him simply buy some cool software, unfortunately. I hope there is a way to use clever Google search operators or similar things to go around this issue. Many thanks in advance!

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  • are there code signing certificates cheaper than US $99 per year? [closed]

    - by gerryLowry
    ksoftware discounts Commodo code signing certificates to US $99 per year. in the past, I've seen Commodo code signing certificates for US $80. I'm excluding CAcert which AFAIK are FREE but are not covered by browsers like IE AFAIK. QUESTION: What is the best price per year for a code signing certificate? Thank you ~~ gerry (lowry) Edit: **THIS SHOULD NOT HAVE BEEN CLOSED** from the FAQ: http://stackoverflow.com/faq ---------------------------- What kind of questions can I ask here? Programming questions, of course! As long as your question is: * detailed and specific <====== YES! * written clearly and simply <====== YES! * of interest to other programmers <====== YES! I've been programming for over 40 years. http://gerrylowryprogrammer.com/ I've taught computer programming at the community college level. I'm a Star level contributer to forums.asp.net. http://forums.asp.net/members/gerrylowry.aspx imo, I've a very good idea what is of interest to other programmers. MORE INFORMATION Ensuring the integrity of code and executables that one distributes is just as much about programming as is knowing how to flip bits in assembler, use delegates in c#, and use the BDD context/specification still of "test first testing".

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  • Django: Filtering datetime field by *only* the year value?

    - by unclaimedbaggage
    Hi folks, I'm trying to spit out a django page which lists all entries by the year they were created. So, for example: 2010: Note 4 Note 5 Note 6 2009: Note 1 Note 2 Note 3 It's proving more difficult than I would have expected. The model from which the data comes is below: class Note(models.Model): business = models.ForeignKey(Business) note = models.TextField() created = models.DateTimeField(auto_now_add=True) updated = models.DateTimeField(auto_now=True) class Meta: db_table = 'client_note' @property def note_year(self): return self.created.strftime('%Y') def __unicode__(self): return '%s' % self.note I've tried a few different ways, but seem to run into hurdles down every path. I'm guessing an effective 'group by' method would do the trick (PostGres DB Backend), but I can't seem to find any Django functionality that supports it. I tried getting individual years from the database but I struggled to find a way of filtering datetime fields by just the year value. Finally, I tried adding the note_year @property but because it's derived, I can't filter those values. Any suggestions for an elegant way to do this? I figure it should be pretty straightforward, but I'm having a heckuva time with it. Any ideas much appreciated.

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  • Should I be worried if I don't get any internships by 3rd year's end?

    - by karamba
    I am in some mediocre college in some corner of India. Am about to complete 3rd year C.S. in a month and a half. I have no idea how to go about "finding an internship" as everyone seems to put it. Looking at online advice, I find that a primary way is to "use your contacts". I am sad to say that I don't have many friends(those I have, I am trying to get help from them for all it's worth), and my family can't help me as they have no idea about the software industry. My college has no official facility for aiding students in this, and the few faculty members who had contacts in "whatever" part of the industry have favoured some students that they have personally come to know. (Though I hear that the "internships" they got involve them stocking equipment in some small companies.... still it's something?) I'm getting nervous. I am considering just spending the coming summer refining my skills in C++ and begin learning MySQL and C#, both of which I have zero experience in. Maybe work on my own project... like a library management system. Relative to those in my college, I think I am among the best programmers there, but that isn't saying much as a lot of students can barely write basic code. I have experience in teaching myself C++, and DirectX9 having created a Tetris clone, some basic 3D apps (bouncing balls), and a basic console-based, text-file-database-using library management system (which I plan to improve this summer). Is it alright if I spend my summer so? Will I be able to get a job later on? I know I have to improve my social skills to get anywhere in life, and I will try, but say I am stuck like this till 4th year's end... will such self studying, online learning help me in landing a decent job? Perhaps after I have learned a bit more, joining some open source project?

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  • Group and sort blog posts by date in Rails

    - by Senthil
    I've searched all over web and have not found the answer. I'm trying to have a very standard archive option for my blog based on date. A request to url blog.com/archive/2009 shows all posts in 2009, blog.com/archive/2009/11 shows all posts in November 2009 etc. I found two different of code but not very helpful to me. def display_by_date year = params[:year] month = params[:month] day = params[:day] day = '0'+day if day && day.size == 1 @day = day if (year && month && day) render(:template => "blog/#{year}/#{month}/#{date}") elsif year render(:template => "blog/#{year}/list") end end def archive year = params[:year] month = params[:month] day = params[:day] day = '0'+day if day && day.size == 1 if (year && month && day) @posts_by_month = Blog.find(:all, :conditions => ["year is?", year]) else @posts_by_month = Blog.find(:all).group_by { |post| post.created_at.strftime("%B") } end end Any help is appreciated.

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  • Explaining NULL and Empty to your 6-year old?

    - by Atomiton
    I'm thinking in terms of Objects here. I think it's important to simplify ideas. If you can explain this to a 6-year old, you can teach new programmers the difference. I'm thinking that a cookie object would be apropos: public class Cookie { public string flavor {get; set; } public int numberOfCrumbs { get; set; } }

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  • How would you explain your job to a 5-year old?

    - by Canavar
    Sometimes it's difficult to define programming to people. Especially too old or too young people can not understand what I do to earn money. They think that I repair computers, or they want to think that I (as an engineer) build computers at work. :) It's really hard to tell people that you produce something they can't touch. Here is a funny question, how would you explain your job to a 5-year-old?

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  • Most concise way to convert from date format: yyyy[3 digit day of year] to SQL datetime

    - by Seth Reno
    I'm working with an existing database where all dates are stored as integers in the following format: yyyy[3 digit day of year]. For example: 2010-01-01 == 2010001 2010-12-31 == 2010356 I'm using the following SQL to convert to a datetime: DATEADD(d, CAST(SUBSTRING( CAST(NEW_BIZ_OBS_DATE AS VARCHAR), 5, LEN(NEW_BIZ_OBS_DATE) - 4 ) AS INT) - 1, CAST('1/1/' + SUBSTRING(CAST(NEW_BIZ_OBS_DATE AS VARCHAR),1,4) AS DATETIME)) Does anyone have a more concise way to do this?

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  • 2010's Most Popular Articles

    The end of the year is upon us, 2010 is about to be in the books. When closing out a year I like to take a look back at the articles I wrote over the year and see which ones resonated the most with readers. Which ones generated the most reader emails? Which ones were read the most? Such a retrospective analysis highlights what content was of most interest to developers in the trenches, and this data is used to guide article topics in the new year. I ended last year with a "Best Of" article - see 2009's Most Popular Articles - and decided to continue this tradition. Such "Best Of" articles give both regular and new readers a chance to discover (or rediscover) the most favored content from the year. So here it is - a list and synopsis of the 2010's most popular articles on 4GuysFromRolla.com. Read More >

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  • Oracle Excellence Awards 2012

    - by A&C Redaktion
    Spezialisierte Partner können sich ab sofort bis 29. Juni als „Specialized Partner of the Year“ bewerben. Damit honoriert Oracle diejenigen OPN Partner in EMEA, die sich mithilfe der Spezialisierung besonders ausgezeichnet haben, sei es durch einen hohen Mehrwert für deren Endkunden oder innovativen Lösungen und Services. Voraussetzung für eine Bewerbung ist mindestens eine abgeschlossene Spezialisierung in diesen sieben Kategorien: Specialized Partner of the Year: Database Specialized Partner of the Year: Applications Specialized Partner of the Year: Middleware Specialized Partner of the Year: Industry Specialized Partner of the Year: Oracle Accelerate Specialized Partner of the Year: Servers and Storage Systems Specialized Partner of the Year: Oracle on Oracle Der Gewinner eines “Specialized Partner of the Year” EMEA Awards erhält 5.000 US-Dollar MDF und vielfältige Möglichkeiten, sich in Szene zu setzen. Wie auch im letzten Jahr wird der Award wieder auf der Oracle OpenWorld in San Francisco überreicht. Interviews, Videos, Werbung, Berichte im Oracle Magazine und ein kostenloser Konferenzpass sind natürlich inklusive. Wie immer, gilt die Bewerbung für den EMEA-Award gleichzeitig als Bewerbung für den deutschen Partner-Award 2012, der auf dem Oracle Partner Day (im Herbst nach der OpenWorld) verliehen wird. Normal 0 21 false false false DE X-NONE X-NONE MicrosoftInternetExplorer4 /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0cm 5.4pt 0cm 5.4pt; mso-para-margin:0cm; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:11.0pt; font-family:"Calibri","sans-serif"; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:"Times New Roman"; mso-fareast-theme-font:minor-fareast; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin; mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi;} Die Nominierung Ihres erfolgreichen Projektes muss diesmal hier auf Englisch eingereicht werden, da eine internationale Jury entscheidet. Beschreiben Sie Ihr Projekt so ausführlich wie möglich, damit sich die Juroren ein gutes Bild Ihrer Leistungen und Services machen können. Wenn Sie hierbei Unterstützung benötigen, fragen Sie einfach Ihren Channel Manager. Denn: Je aussagekräftiger Ihre Unterlagen sind, desto höher ist Ihre Chance zu gewinnen! Alle Informationen zu den diesjährigen Awards finden Sie auf der Oracle Excellence Awards Website. Wir drücken Ihnen die Daumen!

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  • Oracle Excellence Awards 2012

    - by A&C Redaktion
    Spezialisierte Partner können sich ab sofort bis 29. Juni als „Specialized Partner of the Year“ bewerben. Damit honoriert Oracle diejenigen OPN Partner in EMEA, die sich mithilfe der Spezialisierung besonders ausgezeichnet haben, sei es durch einen hohen Mehrwert für deren Endkunden oder innovativen Lösungen und Services. Voraussetzung für eine Bewerbung ist mindestens eine abgeschlossene Spezialisierung in diesen sieben Kategorien: Specialized Partner of the Year: Database Specialized Partner of the Year: Applications Specialized Partner of the Year: Middleware Specialized Partner of the Year: Industry Specialized Partner of the Year: Oracle Accelerate Specialized Partner of the Year: Servers and Storage Systems Specialized Partner of the Year: Oracle on Oracle Der Gewinner eines “Specialized Partner of the Year” EMEA Awards erhält 5.000 US-Dollar MDF und vielfältige Möglichkeiten, sich in Szene zu setzen. Wie auch im letzten Jahr wird der Award wieder auf der Oracle OpenWorld in San Francisco überreicht. Interviews, Videos, Werbung, Berichte im Oracle Magazine und ein kostenloser Konferenzpass sind natürlich inklusive. Wie immer, gilt die Bewerbung für den EMEA-Award gleichzeitig als Bewerbung für den deutschen Partner-Award 2012, der auf dem Oracle Partner Day (im Herbst nach der OpenWorld) verliehen wird. Normal 0 21 false false false DE X-NONE X-NONE MicrosoftInternetExplorer4 /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0cm 5.4pt 0cm 5.4pt; mso-para-margin:0cm; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:11.0pt; font-family:"Calibri","sans-serif"; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:"Times New Roman"; mso-fareast-theme-font:minor-fareast; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin; mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi;} Die Nominierung Ihres erfolgreichen Projektes muss diesmal hier auf Englisch eingereicht werden, da eine internationale Jury entscheidet. Beschreiben Sie Ihr Projekt so ausführlich wie möglich, damit sich die Juroren ein gutes Bild Ihrer Leistungen und Services machen können. Wenn Sie hierbei Unterstützung benötigen, fragen Sie einfach Ihren Channel Manager. Denn: Je aussagekräftiger Ihre Unterlagen sind, desto höher ist Ihre Chance zu gewinnen! Alle Informationen zu den diesjährigen Awards finden Sie auf der Oracle Excellence Awards Website. Wir drücken Ihnen die Daumen!

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  • Do you know of a C macro to compute Unix time and date?

    - by Alexis Wilke
    I'm wondering if someone knows/has a C macro to compute a static Unix time from a hard coded date and time as in: time_t t = UNIX_TIMESTAMP(2012, 5, 10, 9, 26, 13); I'm looking into that because I want to have a numeric static timestamp. This will be done hundred of times throughout the software, each time with a different date, and I want to make sure it is fast because it will run hundreds of times every second. Converting dates that many times would definitively slow down things (i.e. calling mktime() is slower than having a static number compiled in place, right?) [made an update to try to render this paragraph clearer, Nov 23, 2012] Update I want to clarify the question with more information about the process being used. As my server receives requests, for each request, it starts a new process. That process is constantly updated with new plugins and quite often such updates require a database update. Those must be run only once. To know whether an update is necessary, I want to use a Unix date (which is better than using a counter because a counter is much more likely to break once in a while.) The plugins will thus receive an update signal and have their on_update() function called. There I want to do something like this: void some_plugin::on_update(time_t last_update) { if(last_update < UNIX_TIMESTAMP(2010, 3, 22, 20, 9, 26)) { ...run update... } if(last_update < UNIX_TIMESTAMP(2012, 5, 10, 9, 26, 13)) { ...run update... } // as many test as required... } As you can see, if I have to compute the unix timestamp each time, this could represent thousands of calls per process and if you receive 100 hits a second x 1000 calls, you wasted 100,000 calls when you could have had the compiler compute those numbers once at compile time. Putting the value in a static variable is of no interest because this code will run once per process run. Note that the last_update variable changes depending on the website being hit (it comes from the database.) Code Okay, I got the code now: // helper (Days in February) #define _SNAP_UNIX_TIMESTAMP_FDAY(year) \ (((year) % 400) == 0 ? 29LL : \ (((year) % 100) == 0 ? 28LL : \ (((year) % 4) == 0 ? 29LL : \ 28LL))) // helper (Days in the year) #define _SNAP_UNIX_TIMESTAMP_YDAY(year, month, day) \ ( \ /* January */ static_cast<qint64>(day) \ /* February */ + ((month) >= 2 ? 31LL : 0LL) \ /* March */ + ((month) >= 3 ? _SNAP_UNIX_TIMESTAMP_FDAY(year) : 0LL) \ /* April */ + ((month) >= 4 ? 31LL : 0LL) \ /* May */ + ((month) >= 5 ? 30LL : 0LL) \ /* June */ + ((month) >= 6 ? 31LL : 0LL) \ /* July */ + ((month) >= 7 ? 30LL : 0LL) \ /* August */ + ((month) >= 8 ? 31LL : 0LL) \ /* September */+ ((month) >= 9 ? 31LL : 0LL) \ /* October */ + ((month) >= 10 ? 30LL : 0LL) \ /* November */ + ((month) >= 11 ? 31LL : 0LL) \ /* December */ + ((month) >= 12 ? 30LL : 0LL) \ ) #define SNAP_UNIX_TIMESTAMP(year, month, day, hour, minute, second) \ ( /* time */ static_cast<qint64>(second) \ + static_cast<qint64>(minute) * 60LL \ + static_cast<qint64>(hour) * 3600LL \ + /* year day (month + day) */ (_SNAP_UNIX_TIMESTAMP_YDAY(year, month, day) - 1) * 86400LL \ + /* year */ (static_cast<qint64>(year) - 1970LL) * 31536000LL \ + ((static_cast<qint64>(year) - 1969LL) / 4LL) * 86400LL \ - ((static_cast<qint64>(year) - 1901LL) / 100LL) * 86400LL \ + ((static_cast<qint64>(year) - 1601LL) / 400LL) * 86400LL ) WARNING: Do not use these macros to dynamically compute a date. It is SLOWER than mktime(). This being said, if you have a hard coded date, then the compiler will compute the time_t value at compile time. Slower to compile, but faster to execute over and over again.

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