[Bug fortran/30664] -pedantic: "Integer outside symmetric range" for integer(1) and (2) does not work

[fxcoudert at gcc dot gnu dot org]

------- Comment #1 from fxcoudert at gcc dot gnu dot org  2007-02-01 12:36 -------
-huge(i2)-1 is a integer(kind=4) value, because 1 is 1_4 (the default kind) and
 -huge(i2) is thus promoted. You get warnings if you write it with
-huge(i2)-1_2, like in the following code:

program main
implicit none
integer(1) :: i1(3), a1(3:2)
integer(2) :: i2(3), a2(3:2)
integer(4) :: i4(3), a4(3:2)
integer(8) :: i8(3), a8(3:2)
logical :: msk(3)
msk = .false.

i1 = 1
i2 = 1
i4 = 1
i8 = 1

if(-huge(i1)-1_1 /= maxval(i1, msk)) call abort()
if(-huge(a1)-1_1 /= maxval(a1))      call abort()

if(-huge(i2)-1_2 /= maxval(i2, msk)) call abort()
if(-huge(a2)-1_2 /= maxval(a2))      call abort()

if(-huge(i4)-1_4 /= maxval(i4, msk)) call abort()
if(-huge(a4)-1_4 /= maxval(a4))      call abort()

if(-huge(i8)-1_8 /= maxval(i8, msk)) call abort()
if(-huge(a8)-1_8 /= maxval(a8))      call abort()
end program main


-- 

fxcoudert at gcc dot gnu dot org changed:

           What    |Removed                     |Added
----------------------------------------------------------------------------
             Status|UNCONFIRMED                 |RESOLVED
         Resolution|                            |INVALID


http://gcc.gnu.org/bugzilla/show_bug.cgi?id=30664