Dates...

[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

[ustk]

@Lindon the push’s been made, Christoph should merge soon

[d.healey]

@ustk Already has, no?

Commits · christophhart/HISE

The open source framework for sample based instruments - Commits · christophhart/HISE

GitHub (github.com)

[ustk]

@d-healey Oh yes my mistake, was waiting for the others...