subroutine bound(n, x, xl, xu)
integer n, i
double precision x(n), xl(n), xu(n)
do i = 1, n
! Note that xl(i) and xu(i) may be NaN to indicate no bound
if(xl(i).eq.xl(i).and.x(i) < xl(i))then
x(i) = xl(i)
else if(xu(i).eq.xu(i).and.x(i) > xu(i))then
x(i) = xu(i)
end if
end do
end subroutine bound
src/jni/fortran/slsqp.f90
subroutine bound(n, x, xl, xu) integer n, i double precision x(n), xl(n), xu(n) do i = 1, n ! Note that xl(i) and xu(i) may be NaN to indicate no bound if(xl(i).eq.xl(i).and.x(i) < xl(i))then x(i) = xl(i) else if(xu(i).eq.xu(i).and.x(i) > xu(i))then x(i) = xu(i) end if end do end subroutine bound
what does this sentence do？ xl(i).eq.xl(i)