Dates...

[Christoph Hart]

Hmm, Timing is a pretty bad name for this as I would expect it to measure some benchmark stuff or handle MIDI delays etc.

What's the problem with Time? I'm not aware of any existing classes with this name and even if, I would prefer ´Date` as an API class (there's also a popular JS library with this name so it's a bit more close to our mother language.

[Christoph Hart]

reg leapYears = [1992,1996,2000,2004,2008,2012,2016,2020,2024,2028,2032]; //add more if you need em

Lol

inline function isLeapYear(year) 
{
    return year % 4 == 0;
}

[ustk]

@Christoph-Hart Simply because this was as far as my english brought me today

I tried Time but I encountered some conflicts with Juce Time, or I'm just idiot in the way I did it...

Date sounds good :thumbs_up:

[Christoph Hart]

@ustk ah your class doesn‘t need to be called Time in C++ the scripting name is determined by getObjectName().

[ustk]

@Christoph-Hart If you say so :man_shrugging: Could you say it again in french please?

EDIT: Oh ok I see :thumbs_up:

[d.healey]

Yeah I think date is a good choice since that's familiar to us Javascript people :)

[ustk]

@Christoph-Hart @d-healey @Lindon Pull request made (along with a few others)

[Lindon]

@Christoph-Hart said in Dates...:

reg leapYears = [1992,1996,2000,2004,2008,2012,2016,2020,2024,2028,2032]; //add more if you need em

Lol

inline function isLeapYear(year) 
{
    return year % 4 == 0;
}

except of course this is wrong.... you will need

year % 100

and

year % 400

[Christoph Hart]

@Lindon huh? This would mean that theres a leap year every 400 years so please tell everyone who has their birthday on the 29th of February that they won‘t live to see another birthday party…

[Lindon]

@Christoph-Hart - er you should go read up on what makes a leap year, here's the code (which I couldnt be bothered writing and testing last time out - naturally lazy...):

inline function isLeapYear(year) 
{
    if(year % 400 == 0 )
    {
           return true;
    }else{
        if(year % 100 == 0)
        {
             return false;
        }else{
            return year % 4 == 0;
       }
    }
}

So 2000 is a leap year - but 1900 isnt.

[Dan Korneff]

@Lindon here's a little script online:
https://www.programiz.com/javascript/examples/check-leap-year

inline function isLeapYear(year) 
{
     if ((0 == year % 4) && (0 != year % 100) || (0 == year % 400)) 
     {
          return true;
     }  
     else 
     {
          return false;
     }
}

[d.healey]

@Dan-Korneff Why not?

inline function isLeapYear(year) 
{
    return (0 == year % 4) && (0 != year % 100) || (0 == year % 400));
}

[Dan Korneff]

@d-healey That works too, and you save 4 lines of code.

[ustk]

[Lindon]

@Dan-Korneff said in Dates...:

@d-healey That works too, and you save 4 lines of code.

if we are interested in saving lines of code my original version is precisely 1 line.... :-)

[Dan Korneff]

@Lindon I only skimmed the last few posts

[Christoph Hart]

if we are interested in saving lines of code my original version is precisely 1 line....

But you need to ship an update in 2036 and we're all in here for the long run...

[Lindon]

@Christoph-Hart said in Dates...:

if we are interested in saving lines of code my original version is precisely 1 line....

But you need to ship an update in 2036 and we're all in here for the long run...

well you have me there...

[Lindon]

@d-healey said in Dates...:

@Dan-Korneff Why not?

inline function isLeapYear(year) 
{
    return (0 == year % 4) && (0 != year % 100) || (0 == year % 400));
}

except when you try this it always returns true....so it doenst work.

-- oh hang on I spot a typeo in my implementation....

[Lindon]

Okay so whilst we await @ustk 's push here's a fix using just HISE:

function isLeapYear(aYear)
{
	return ((0 == aYear % 4) && (0 != aYear % 100) || (0 == aYear % 400));
}

function daysFrom2000(aYear,aMonth,aDay)
{
	reg calcYear = 2000;
	reg totalDays = 0;
	reg leap = 0;
	reg monthAmounts = [31,28,31,30,31,30,31,31,30,31,30,31];
	if(aYear < calcYear)
		return -1;
		
	if(aMonth > 2 && isLeapYear(aYear))
		leap = 1;
	totalDays = aDay + leap;
	for(i = 0; i< aMonth-1; i++)
	{
		totalDays = totalDays + monthAmounts[i];
	};
		
		
	while(calcYear < aYear)
	{
		totalDays = totalDays + 365 + isLeapYear(calcYear);
		calcYear++;
	};
	
	return totalDays;
}

Console.print(daysFrom2000(2023,01,01));

Obviously just doing days not microseconds