Rounding Numbers / Damage Dealing

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AuthorTopic: Rounding Numbers / Damage Dealing
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How does one round numbers? For instance, a value divided by two. And if that's impossible, can damage be dealt in increments of .5?

(EDIT again: Hah! I found a use for this topic after all. ^_^)

[ Monday, July 19, 2004 10:55: Message edited by: Sathoj ]

Posts: 6936 | Registered: Tuesday, September 18 2001 07:00
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(untested pseudo-code)
if (my_parm % 2 > 5) { // div remainder greater than 5
my_parm = (my_parm / 2) + 1; // round it up

Posts: 200 | Registered: Wednesday, March 31 2004 08:00
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I believe decimals get rounded down. BoA just cuts off the part after the decimal point.

if (25/6 == 29/6)
print_str("It's true!");
I think I tried this once and it printed

Debug value: 4
Debug value: 4
It's true!

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Posts: 7968 | Registered: Saturday, February 28 2004 08:00
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Originally written by Just call me Kel:

I believe decimals get rounded down. BoA just cuts off the part after the decimal point.
Right. That's why if the remainder (x modulo y returns the remainder from integer division) is more than 5 (which means more than .5), you can just do the integer div and add 1 to "round up". :)

EDIT: That's what I get for posting before my "morning" coffee (I keep odd hours :) ). The pseudo-code should be:

if ((x % y) >= (y / 2))
x = (x / y) + 1;
x = x / y;
Obviously, the remainder won't always be greater than '5', but instead half or more of the divisor. :D

[ Monday, July 19, 2004 12:28: Message edited by: spyderbytes ]

Posts: 200 | Registered: Wednesday, March 31 2004 08:00
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Uhhh, wouldn't it be easier just to do this:

y = y * 10;
if ((y / 2) % 10 == 5)
EDIT: I believe this is the same behavior as C and Java, isn't it? When you declare int variables and divide, it just chops off the stuff after the decimal point.

I think this because I vaguely remember doing 5/2=2 in Java.

[ Tuesday, July 20, 2004 07:04: Message edited by: Keep ]

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Posts: 1415 | Registered: Thursday, March 27 2003 08:00
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Not sure why that would be easier. :)

The pseudo-code I came up with once I finally pulled my head out of the dark cavity it was in has the advantage of being able to round up regardless of the divisor (and yes, the original post did mention "like dividing by 2", but bulletproof is always better :) ).

Also, seems to me that if all you're trying to determine is what's even and what's odd, it would be easier to just:

if ((x % 2) == 0)
...I'm even...
...I'm odd...
Guess I just missed the point of multiplying by 10 then dividing by 2 and checking for a 5 remainder modulo 10... but it's pre-coffee time for me again. :D

Posts: 200 | Registered: Wednesday, March 31 2004 08:00