------- Comment #11 from kargl at gcc dot gnu dot org 2010-05-31 21:54 ------- (In reply to comment #7) > (In reply to comment #6) > > because in the above 'i' and 'I' are in the same scoping unit. > I couldn't find what mandates in the standard that i and I are in a different > scoping unit/are different variables. Are they ? > > > If you write 'i = 5; j = [(i,i=1,I)]' then the 'i' here is in > > a different scoping unit. I agree that the scalar-int-expr > > '1' and 'I' need to be evaluated to establish the the loop > > start and stop values. The question again based on scoping > > unit is whether 'I' is uninitialized. > > > How could it be ? >
I don't understand what you are asking. integer j(5) I = 5 j = (/ (i,i=1,I) /) ! (i,i=m1,m2) for discussion below end 'I' in line 2 is in the scope of the main program. 'i' in line 3 is in the scope of the implied-do-loop. 'I' in line 2 is not the same as 'i' in line 3. The only thing that 'i' in line 3 gets from 'I' in line 2 is its type and kind type parameter. When 'i' in line 3 comes into scope, does 'I' in the m2 become undefined? I can't find anything in the standard that states that it becomes undefined and I can't find anywhere that states that it is still defined to have a value of 5 in the evaluation of m2. -- http://gcc.gnu.org/bugzilla/show_bug.cgi?id=44354
