How to decide if a sprite is close enough?

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Charles Black

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Posted: 6th Nov 2008 23:34

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I'm making a tower defense and was wondering how you could tell when the enemy sprite is close enough to the tower's range for it to attack.

Please explain what I have to do, don't just give me the solution straight away.

Charles

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Zuka

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Posted: 7th Nov 2008 00:19

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You could use (maybe) a quicker method, calculating the tiles the tower can fire, and if the enemy is in that tile, fire at it...

or, the much easier method - checking distance using the Pythagorean theorem.

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Charles Black

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Posted: 7th Nov 2008 17:21

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lol

how does the Pythagorean theorem work?

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prasoc

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Posted: 7th Nov 2008 21:43

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to put it (kinda) simply, Pythagoras stated that you can work out any sides in a right angle triangle if you have 2 of the 3 sizes. so to work out the distance how far away the object is, you minus the turret x co-ord from the enemy co-ord (dont forget to "abs()" it) and the turret y co-ord from the enemy y co-ord and then use Pythagoras' theorem to work out the size of the third side (hypotenuse) and check to see if it is under a certain size

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Zuka

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Posted: 7th Nov 2008 23:11

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Or...

x3 = dbABS(x1 - x2);
y3 = dbABS(y1 - y2);

distance = dbSQRT(x3 * x3 + y3 * y3 + z3 * z3);

Most people would put ()'s around the squaring, but that's useless, seeing as C++ sticks to the PEMDAS (order of operations) rule.

The sqrt() in the header "math.h" is faster than the GDK's.

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Charles Black

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Posted: 8th Nov 2008 23:37

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where does the z3 come from?

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jezza

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Posted: 9th Nov 2008 09:27

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take out the + z3 * z3 at the end.

p.s. you call it PEMDAS? We were taught BIDMAS.
brackets
indices
div/mult
add/sub

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Charles Black

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Posted: 9th Nov 2008 13:15

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And, what does dbABS do, exactly?
this is the abs() function I found through google:

+ Code Snippet

 Returns the absolute value.
Example: 

/* abs example */
#include <stdio.h>
#include <stdlib.h>

int main ()
{
  int n,m;
  n=abs(23);
  m=abs(-11);
  printf ("n=%d\n",n);
  printf ("m=%d\n",m);
  return 0;
}

Output: 

n=23
m=11

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jezza

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Posted: 9th Nov 2008 20:25

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an absolute value is a positive one. eg -5 becomes 5, -10 becomes 10 but 3 stays as 3

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SunDawg

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Posted: 10th Nov 2008 00:46 Edited at: 10th Nov 2008 00:47

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If you include math.h, you can do all of that more quickly.

+ Code Snippet

distance=sqrt(pow(towerPositionX-enemyPositionX,2),pow(towerPositionY-enemyPositionY,2));

That would give you the distance.

As for an absolute value, it's technically a value's distance from zero, which is therefore always positive. Just as speed, a distance lacks direction; They're scalar values. Unlike vector values (displacement, velocity, &c.) which have magnitude and direction, and can therefore be negative.