Fortran overflow algorithm

Question

I need to know, what is the problem with my overflow algorithm?, In Fortran I have to do an algorithm to return the number of the increment "n" when it reach the overflow break.

program overflow

  integer::n,i,fact;

  fact = 1;
  n = 50;

  do i=1,n,1
     fact=fact*i;
     if ((fact==abs(fact)).and.(fact /= 0)) then
        print*,"F!=",fact;
        n=n+1;
        !print*,"Overflow=",n;
     end if;
  end do;   

  print*,n;

end program overflow

Its contains negative values, and "mutations" in factorial before reach the number that means "overflow", but it's false, n should be 15.

possible duplicate of [Catch integer exceptions in Fortran](http://stackoverflow.com/questions/18881865/catch-integer-exceptions-in-fortran) — Alexander Vogt, Oct 19 '13 at 07:24
interesting, we know you cant *count on* an overflow going negative, but shouldn't this code catch it when it does? — agentp, Oct 19 '13 at 14:15

score 0 · Accepted Answer · answered Oct 19 '13 at 02:03

0

I have never been a Fortran programmer but perhaps your return value should be a double.

answered Oct 19 '13 at 02:03

Ross Bush

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Fortran overflow algorithm

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