math question (beginner)

The compiler will do the conversion for you when you do an assignment. Just assign a float to an integer.

For rounding, the easiest thing to do is always add 0.5 to the float. It will take more time and code space to check whether you need to do so than it does just to add 0.5 and have it truncated away again.

i = f; // truncate
i = f + 0.5; // round

At some point you might be interested in the floor() and ceil() functions in math.h.

The compiler will do the conversion for you when you do an assignment. Just assign a float to an integer.

For rounding, the easiest thing to do is always add 0.5 to the float. It will take more time and code space to check whether you need to do so than it does just to add 0.5 and have it truncated away again.

i = f; // truncate
i = f + 0.5; // round

At some point you might be interested in the floor() and ceil() functions in math.h.

I would have guessed that execution would be faster to cast to an int then add 1 as opposed to doing a floating point addition (1 in Nocky's example and .5 in Drew's) then casting to an int.

Then again I could be wrong, you could always try it both ways and find out.

Walt

Surely casting to int then adding one will always round up?

Stefan

I was thinking of something like this but I see there is still one floating point math operation. Unless someone can think of a way to eliminate it, Drew's way might be best.

int Convertor(float fSrc)
{
   if ((fSrc - (int)fsrc) >= 0.5)
      return ((int)fSrc+1)); // instead of return((int)(fSrc+1));

   else
      return( (int)fSrc);
}

You've actually got two floating point operations - the subtraction and the comparison.

I think Drew's method is definitely the cheapest and simplest.

Stefan

"I think Drew's method is definitely the cheapest and simplest."

Yes, for positive numbers only. For negative numbers, you'd want to subtract 0.5 (add -0.5).