Float -> Int

[legacy-^::B!G-A::]

Ok im wondering if i do this, Int(MYFLOAT) does it Round the number to an int like Round() does or does it just remove the decimal? so if its 1.2 or 1.9 it returns 1?

[Xyx]

int() simply chops off the decimal part, so 1.9 would become 1.

Here's an easy and documented way for you to check such things in the future; a commandlet.

//====================================================================
// Generic Commandlet.
//
// Use as follows:
//  - Open a DOS box.
//  - Navigate to the UT2004\System folder.
//  - Enter the following command:
//
//    UCC [package name].[commandlet name] [parameter]...
//
//    package name:
//    The name of the package containing the commandlet.
//
//    commandlet name:
//    The name of your commandlet class.  If the class name ends in
//    "commandlet", you may leave that out.  The engine will add it.
//
//    parameter:
//    Optional string parameter.
//====================================================================

class XS_Commandlet extends Commandlet;

event int Main(string Parameter)
{
    // Whitespace for readability;
    Log("");

    // Replace this line with whatever you wish to display in the DOS 
    // box.
    Log("Testing " $ Parameter);

    return 0;
}

/** Concatenates v to a string in (X,Y,Z) format. */
static final operator(40) string $ (coerce string s, vector v)
{
    local string NewString;

    NewString = s $ Round(v.X) $ "," $ Round(v.Y) $ "," $ Round(v.Z);

    return NewString;
}

/** Concatenates r to a string in (X,Y,Z) format. */
static final operator(40) string $ (coerce string s, rotator r)
{
    local string NewString;

    NewString = s $ r.Pitch $ "," $ r.Yaw $ "," $ r.Roll;

    return NewString;
}

I actually used this to confirm the result.

[legacy-^::B!G-A::]

Thx a bunch but how do i use that c.l. to do anything? lol like tell me how u solve my question

[legacy-ghouck]

Of course, if you want to be REALLY cheap, you could try
int(MyFloat + 0.5), which gives you the equivalent of "rounding off"

[legacy-^::B!G-A::]

well thats not wat i want, i want it so no matter what the Decimal is and the integer is 1 it will always return 1, 1.12058021 or 1.01 or 1.98 will come back 1, its for my award stats check:

function bool CheckAward(int Param1, int Param2)
{
   if( Param1/Param2 == Int(Param1/Param2) )
      return true;
   else
       return false;
}

[Xyx]

Put the commandlet in whatever package you like, replace the line marked with the "replace this line" comment with Log(int(1.9)); and compile it and follow the instructions in the comments at the top of the code. That's all there is to it.

[legacy-^::B!G-A::]

oh lol "Biga: Biga your a dur. Biga: I KNO! THX FOR NOTICING"

[legacy-^::B!G-A::]

I seem to be having a problem.

function bool CheckAward(int Param1, int Param2)
{
   local float f;
   f = Param1/Param2;
   if( f == Int(f) )
      return true;
   else
       return false;
}

always returns true when Param1=1 and Param2=12?

so wouldnt it be:

function bool CheckAward(int Param1, int Param2)
{
   local float f;
   f = 1/12; // should be 0.0833 (repeated)
   if( f == Int(f) ) // if( 0.0833 == 0 ) should be false right?
      return true;
   else
       return false;
}

EDIT:

i ran a test with logs:
ScriptLog: Param1=1 Param2=12
ScriptLog: f=0.00
ScriptLog: if( 0.00 == 0 )

it seems i need the float to be longer then 2 decimal places, more then .00 i need more like .000000

[legacy-^::B!G-A::]

OK i fixed it, i timed Param1 and Param2 by 10 and it worked, im going to items by 100 just to be sure i passes over a int that will dived to 0

[Xyx]

I think the engine makes an integer division because you're dividing two integers, and only then decides to make it a float because you're storing the result in a float. You could probably get the engine to make a float right away like this: float(Param1) / Param2. Then it would no longer be an integer division.

[legacy-^::B!G-A::]

im sry lol i didnt understand that:noob:

[Xyx]

I'm not 100% sure, but I think this is how it works:

You try to divide two whole numbers, and the engine gives you a whole number as the result. You try to divide a whole number and a decimal number or two decimal numbrs, the engine gives you a decimal number as the result.

You want to divide two whole numbers, but still get a decimal number as the result. What should you do? Turn one of the whole numbers into a decimal number. You do that in the same way you turned the float into the int earlier on, only the other way around: float(Param1). Now it is treated as a decimal number, and when you divide it by Param2 you will get a decimal number as the result.

[legacy-Bonehed316]

INT/INT = INT

ALWAYS.

1/12 = 0.whatever. Everything after the . gets CHOPPED OFF because its INT/INT. Then the result gets casted to a float, 0.000000.

What you want is float(INT)/INT = float or float(INT)/float(INT) = float. The first way should be faster, and no loss of accuracy.

[legacy-^::B!G-A::]

what i found was, if you divide two ints, it becomes the max of 0.00 but if you divide floats its 0.000000 so i timed by 100 because the 0.000000 still exists but only giving 0.00 so i timed it by 100 to MAKE it show