Finding the “discrete” difference between close floating point numbers

Question

Suppose I have two floating point numbers, x and y, with their values being very close.

There\'s a discrete number of floating point number

Answer 1

Floats are lexicographically ordered, therefore:

int steps(float a, float b){

  int ai = *(int*)&a;  // reinterpret as integer
  int bi = *(int*)&b;  // reinterpret as integer
  return bi - ai;
}

steps(5.0e-1, 5.0000054e-1);  // returns 9

Such a technique is used when comparing floating point numbers.