GustavXIII


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Posted
20110810 19:35:52 » 

Hi! I want to draw just the first Number from Integer. If I have the number 12009, I want to draw only the 1.
In C there was something like number%d or something like that. How does it work for java? It is because I want to make a Timer for the game.




namrog84


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Reply #1  Posted
20110810 19:40:08 » 

in java the % is called a modulus so lets say you have 13%10 that would equal 3 or 11%10 would equal 1, it gives the remainder after the division. this will occur every time timer is divisible by 10, so 1,2,3,4,5,6,7,8,9 would NOT trigger it 1 2 3 4 5
 if(timer % 10 == 0){ } timer++; 
This is 1 of many way some people do some sort of interval timer. You could also just set up a simple little loop to break each number into something separate How exactly do you want to implement your timer? is it forever counting up like elapsed time? or firing/action interval? or whatever, as there may be a better solution for your given scenario. \/ what he said works too \/

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Riven


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Reply #2  Posted
20110810 19:42:47 » 

he wants the leftmost digit: a quick and dirty solution would be: 1
 int digit = String.valueOf(Math.abs(i)).charAt(0)'0'; 
how this relates to timers is a mystery to me.

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GustavXIII


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Reply #3  Posted
20110810 20:38:21 » 

Ok thanks! @namrog In Slick I using delta. If delta is 1000 its 1 second. But I cant show 1000 in the menu so I have to reduce it to the first digit(?).




Gudradain


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Reply #4  Posted
20110810 20:44:36 » 

Divide by 1000




Riven


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Reply #5  Posted
20110810 20:53:08 » 

If I have the number 12009, I want to draw only the 1.
dividing by 1000 won't result in the example the OP posted.

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delt0r


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Reply #6  Posted
20110812 12:23:27 » 

This would also work. 1 2 3 4 5 6
 import static Math.*;
public int firstDig(double x){ double divisor=pow(10,floor(log10(x))); return (int)(x/divisor); } 
This can be extended easily for any number base.

I have no special talents. I am only passionately curious.Albert Einstein



pitbuller


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Reply #7  Posted
20110812 12:28:29 » 

If I have the number 12009, I want to draw only the 1.
dividing by 1000 won't result in the example the OP posted. I think its pretty clear that op ask wrong question. I think he want seconds from milliseconds and somehow messed question.




Addictman


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Reply #8  Posted
20110812 12:39:06 » 

int a = 12009; int b = (int)(a / Math.pow(10, (int)Math.log10(a))); // b > 1
Math.abs() if a is negative.
Edit: Ok, just saw delt0r posted the same.




JL235


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Reply #9  Posted
20110812 14:31:33 » 

Personally, I would go with Rivens String version, but here is an overengineered alternative: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
 public void firstInt( int n ) { boolean positive = false; if ( n < 0 ) { n = Math.abs(n); positive = true; }
for ( int base = 10; base < 1000000000; base *= 10 ) { if ( n < base ) { int result = (int) ( n / (base/10.0) ) ; return positive ? result : result ; } }
throw IllegalStateException( "should never be able to reach this point" ); } 
It could also be expanded into a long list of if's.




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Riven


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Reply #10  Posted
20110812 15:22:21 » 

faster faster! 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
 public static int firstDigit(int n) { int mul;
if (n < 0) { mul = 1; n = n; } else { mul = +1; }
for (int base = 10; base <= 1000000000; base *= 10) { if (n < base) { return mul * n * 10 / base; } }
throw new IllegalStateException("should never be able to reach this point"); } } 

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Roquen


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Reply #11  Posted
20110813 10:00:23 » 

If you really wanted to go overboard you could convert the linear search into a binary one.




counterp


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Reply #12  Posted
20110813 10:06:44 » 

only the original one Riven posted works properly although addictman suggested using Math.abs for negatives so I guess he's good too




delt0r


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Reply #13  Posted
20110813 11:56:46 » 

It won't if the argument is negative. You get "" or whatever negative convention the local uses. I tested my stuff in R seem to work fine.

I have no special talents. I am only passionately curious.Albert Einstein



Riven


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Reply #14  Posted
20110813 14:13:36 » 

If you really wanted to go overboard you could convert the linear search into a binary one.
While this is true, I think it's likely that in this case, most values will be less than 100.000, in which case it's pretty hard to beat the linear search.

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JL235


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Reply #15  Posted
20110813 16:03:26 » 

If you really wanted to go overboard you could convert the linear search into a binary one.
While this is true, I think it's likely that in this case, most values will be less than 100.000, in which case it's pretty hard to beat the linear search. You could increment the base by 100 instead of 10, something like... 1 2 3 4 5 6 7 8 9
 for (int base = 10; base <= 1000000000; base *= 100) { if (n < base) { if ( n < base/10 ) { return mul * n * 10 / (base/10); } else { return mul * n * 10 / base; } } } 
This would almost half the number checks.




Roquen


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Reply #16  Posted
20110813 21:24:52 » 

I thought we were overengineering the problem. Here's an untested entry: 1 2 3 4 5 6
 public static int firstDecimalDigit(int n) { n = Math.abs(n); return (int)Math.floor(n*Math.pow(10, ((int)Math.log10(n)))); } 
edit: I somehow missed the two previous in the same vein...dang!




Abuse


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Reply #17  Posted
20110813 21:55:16 » 

I'm sure you're all aware that negation does not work for Integer.MIN_VALUE (and consequently neither does Math.abs(int)).
So all of the above implementations will malfunction in all kinds of interesting ways when the input n=Integer.MIN_VALUE.




counterp


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Reply #18  Posted
20110813 22:01:56 » 

If you really wanted to go overboard you could convert the linear search into a binary one.
While this is true, I think it's likely that in this case, most values will be less than 100.000, in which case it's pretty hard to beat the linear search. Well in that case 1 2 3 4 5 6 7 8 9
 private static final int digits[] = new int[200001]; static { for (int i = 0; i < digits.length; i++) digits[i] = String.valueOf(Math.abs(i  100000)).charAt(0)'0'; }
public static int lookup(int n) { return digits[n + 100000]; } 
lololololololol 100000 to 100000 if you were worried about memory you could convert it to byte or something, or even do some bitshifting stuff to get maximum value out of the integers




Roquen


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Reply #19  Posted
20110813 22:32:19 » 

Well if you're doing anything like the stuff posted in the thread, then it's almost certain that: You're doing it wrong^{tm}.




delt0r


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Reply #20  Posted
20110814 00:00:00 » 

So Roquen, what is the right way to be doing it?

I have no special talents. I am only passionately curious.Albert Einstein



Riven


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Reply #21  Posted
20110814 00:04:39 » 

I'm fairly certain he means that the 'problem' defined by the OP has some flaws, so the solution will be inherently wrong.

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Roquen


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Reply #22  Posted
20110814 00:11:41 » 

Yes, that's exactly what I meant.




