Math problem - Calculating if an object is in range of another

[Rocco]

i had this several times and dont know a convenient solution.

i want to know if an object is in range within say 10px of another one, and have to write the complicate way like:

if((object[0].X - object[[1].X) < 10 && (object[0].X - object[[1].X) > -10)
do stuff

this construct works, but i'm knowing there is an easier way for this,
cause i have to calculate these several times in my script.

[Calin Leafshade]

Pythagoras!

in pseudo:
distance = sqrt ( (x1 - x2) ^ 2 + (y1 - y2) ^ 2 )

EDIT: the syntax for squaring is not the same in AGS so just multiply the term by itself to square it

[Rocco]

well thx, this will work, but it needs nearly the same effort as before, and i guess it costs more cpu_power.
i thaught in another direction, lets say

pseudo:
distance = (x1 - x2);
if(distance < 0)
make distance positive (exp. convert -13 to 13)

[Calin Leafshade]

your original calculation isnt really that accurate since it uses a kind of 'box' to calculate distance. Mine uses a circual radius around a point.

And on the subject of CPU, unless you are doing this thousands and thousands of times a second it really wont be an issue.

[Rocco]

thats right, i will using your suggestion, big thx. 

btw:
for my interest, maybe you know how to make a value positve?

if(distance < 0)
make distance positive (exp. convert -13 to 13)

[Calin Leafshade]

Code: ags

function AbsInt(int value){

if (value < 0) return value * (-1);
else return value;

}

EDIT: functionised

[Kweepa]

If you are just interested in the linear distance as your code suggests, here is the simplest way.

Code: ags

bool InRange(Object *o1, Object *o2, int range)
{
   int offset = o1.X - o2.X;
   // squaring both sides makes them positive
   return (offset*offset < range*range);
}

if (InRange(o1, o2, 10)
{
   // do stuff
}

[monkey0506]

Calin, you don't have to actually multiply by -1, you can just do this (with the same result, but simpler syntax).

Code: ags

if (distance < 0) distance = -distance;

[Calin Leafshade]

If you are just interested in the linear distance as your code suggests, here is the simplest way.

I completely didnt notice that the y value wasnt a factor.

[Rocco]

big thx, that solutions suits my needs completly.
in this case i also need the y.value,
but the linear approach might come also in handy soon, cause i stumbled often over this kind of calculations in my projects.