http://gcc.gnu.org/bugzilla/show_bug.cgi?id=54934
kargl at gcc dot gnu.org changed:
What |Removed |Added
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CC| |kargl at gcc dot gnu.org
--- Comment #1 from kargl at gcc dot gnu.org 2012-10-15 15:40:28 UTC ---
Jan,
It's a convoluted problem. In gfortran, to get
the maximum range of an integer type you need
to use [-huge(i)-1:huge(i)] where 'i' is of the
type of interest. This is a result of gfortran
seeing a number like -4294967297 as a unary minus
with an operand of 4294967297. 4294967297 without
a kind suffix is of default integer kind, which is
a 32-bit signed int, and 4294967297 is slightly
larger than huge(1) = 2147483647. Note, Fortran
does not have unsigned integers. If you don't
want to or cannot use the -fno-range-check option,
you can do
integer(8), parameter :: n = 2 * (huge(1_4) + 1_8) + 1
dimension foo(n-1:n)
dimension bar(-n-1:-n)
bar = 42
foo = bar
print *, n
stop
end
where I correctly initialize bar to some value; otherwise, the
code is invalid Fortran. You could also decorate the integers
with a integer kind suffix
dimension foo(4294967296_8:4294967297_8)
dimension bar(-4294967297_8:-4294967296_8)
bar = 42
foo=bar
stop
end
Here, the _8 tells gfortran to use a signed 64-bit int.
Both code snippets above compile on my x86_64-*-freebsd.
I don't know want happens on an i686 target.
