Re: Fortran loops

[Adam Hirst <ahirst@cedint.upm.es>]

I don't know how the compiler handles this exactly, but my "educated
guess" would be that it depends on whether the loop is unrolled and on
whether the incrementing of the loop index variable occurs before or
after the check at the "start" of each loop iteration.

If "at" (i.e. just before, while still inside) the end of the "last"
iteration the value is huge(x), and it then decides to add one (causing
it to overflow to a -ve number), and THEN compare to huge(x) to see
whether the loop is over, it ought then go on forever infinitely
repeating the entire span of the integer type.

But I'm not an "informed" person. :)

~ Adam

On 27/06/16 21:36, Dominique d'Humières wrote:
> IIRC this has already been discussed in the past (gfortran.dg/do_1.f90). 
> The Fortran standard requires that the variable i is equal to array(2)+1 on the loop exit, which causes an overflow if array(2) is equal to huge(i). 
> So the code is invalid when b(2) = huge(x), but I don’t see why the overflow of the exit value should lead to an infinite loop.
> Dominique
>> On Jun 27, 2016, at 6:56 PM, Jan Hubicka <hubicka@ucw.cz> wrote:
>>
>> Hello,
>> I have a Fortran question.  The following program loops infinitly when compiled
>> with ifort -O0, while it does the expected number of iterations with gfortran.
>> The issue is that we do produce quite ineffective code to deal with
>> side cases like this.
>>
>> is the program bellow valid and expected to terminate?  If so I guess we want
>> to have it as a testcase.
>>
>> Thanks,
>> Honza
>>
>> program test_program
>> integer(4) :: b(2), a(22), x
>> integer(8) :: suma
>>
>> b(1) = huge(x)-10
>> b(2) = huge(x)
>>
>> call test2(b, suma)
>> print *,suma
>>
>> end program test_program
>> function test2(array, s)
>>  integer(4) :: i, block(9), array(2)
>>  integer (8) :: s
>>  s = 1
>>
>>  do i = array(1), array(2)
>>      s = s + 1
>>  end do
>>
>> end function test2
>>
> 

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