Why is present() not resolved at runtime in Fortran?

Consider the function

integer function add(x,y)
  integer, intent(in) :: x
  integer, intent(in), optional :: y
  
  if (present(y)) then
    add = x+y
  else
    add = x
  endif
endfunction add

I looked at this in Compiler Explorer and it shows that present() is evaluated at runtime with a CMP instruction. Why is this not resolved at compile time? The compiler certainly knows whether the function is called with one or two arguments.