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arjenmarkus wrote:
I do not quite agree: I resolve the unlimited polymorphism using the select type construct. So within these type cases the compiler can and should choose the intrinsic operations, as it knows the actual type of the polymorphic variables. ..
I guess this should come down to the standard - does it say explicitly say as such and if so, where? I don't recall seeing it as such, so it'll be helpful to know it. From what I remember, don't the rules for procedure overloading as described in the standard effectively come down to the defined operation takes precedence over the intrinsic one? Below is a simpler test case using which Intel development team can provide feedback on how the compiler should work in strict accordance with the standard and if there are aspects left up to the processor, how they want to approach it.
module m
implicit none
private
interface operator(<)
module procedure lower
end interface operator(<)
public :: operator(<)
contains
function lower( op1, op2 ) result(islessthan)
!.. argument list
class(*), intent(in) :: op1, op2
!.. function result
logical :: islessthan
!.. local variables
integer :: in1
integer :: in2
print *, " enter lower: "
islessthan = .false.
select type (op1)
type is (integer)
in1 = op1
print *, " op 1 = ", in1
select type (op2)
type is (integer)
in2 = op2
print *, " op 2 = ", in2
islessthan = op1 < op2
end select
type is (real)
select type (op2)
type is (real)
islessthan = op1 < op2
end select
class default
!.. appropriate action elided
end select
end function lower
end module m
program p use m, only : operator(<) implicit none logical :: islessthan islessthan = ( 0 < 42 ) print *, " is 0 < 42? ", islessthan stop end program p
Upon execution,
op 1 = 0 op 2 = 42 enter lower: op 1 = 0 op 2 = 42 enter lower: op 1 = 0 op 2 = 42 enter lower: op 1 = 0 op 2 = 42 enter lower: op 1 = 0 op 2 = 42 enter lower: op 1 = 0 op 2 = 42 enter lower: op 1 = 0 op 2 = 42 enter lower: op 1 = 0 op 2 = 42 enter lower: ...
