Mathf.Lerp for integers

Question

Unexpectedly for myself, I'm hung up on this issue. Suppose there is an array with a length of N elements. And we need to take an element from there, but not by index, but by value t = [0..1] (from 0 to 1 both inclusive). That is, the problem is reduced to a linear interpolation of the float value into the index of the array.

int index = (int) Mathf.Lerp(0, N, t);

But when t = 1, then index = N which is out of array's range.

We can fix this as follows:

int index = Mathf.Clamp((int) Mathf.Lerp(0, N, t), 0, N - 1);

or int index = (int) Mathf.Lerp(0, N, t - Mathf.Epsilon);

but i dont like the way it looks.

So the question is how to do this gracefully?

Answer 1

Best Answer

Answer by Hellium · Jul 18, 2018 at 02:31 PM

After our discussion in the comments, here is the formula that should work:

   // Scaling by ( 1 - Mathf.Epsilon ) prevents N to be chosen when t == 1
  int index = Mathf.FloorToInt( t * N * ( 1 - Mathf.Epsilon ) ) ;

https://www.desmos.com/calculator/5rcgfpnkox

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Kos-Boss · Jul 18, 2018 at 02:40 PM 0

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And now here is exactly the problem that I described initially!

When t = 1, then index = N which is out of array's range. alt text

P. S. You have one redundant closing parentheses in your answer.

Hellium Kos-Boss · Jul 18, 2018 at 03:08 PM 0

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Oh yes, you are right! I guess, you have no choice but scaling by ( 1 - $$anonymous$$athf.Epsilon ) :

 int index = $$anonymous$$athf.FloorToInt( t * N * ( 1 - $$anonymous$$athf.Epsilon ) ) ;

Kos-Boss Hellium · Jul 18, 2018 at 03:14 PM 0

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I hoped there was a way to solve it more elegantly :)

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Mathf.Lerp for integers

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