Dates...

[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?

https://github.com/christophhart/HISE/commits/develop

[ustk]

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