Hi Alan,

the way to get the desired bit pattern is (for instance):

integer, parameter :: PL_FCI_MARK = ishft(1,31)

It shift the single bit of 1 into the 32nd position leaving
a trail of zero bits. (It should even work on ones-complement
machines, though I do not think I have ever been close to
one and knew about it.)

The reason for the failure is that z'80000000' is evaluated
as an 8-bytes integer and therefore you get 2147483648 as
the value. That does not fit into a 4-bytes integer.

Regards,

Arjen

On 2011-01-27 11:11, Alan W. Irwin wrote:
> On 2011-01-27 01:20-0800 Alan W. Irwin wrote:
> 
>> Hi Arjen:
>>
>> I have the sed script almost working for f95.  The only problematic
>> statement generated is
>>
>>       integer, parameter :: PL_FCI_MARK = z'80000000'
>>
>> That z'80000000' causes an integer overflow.  The problem is that
>> z'80000000' is not a normal 4-byte integer.  Assuming twos-complement
>> integers, then the negative of that number is also z'80000000'.  That
>> is, it is a "negative" zero or zero with the sign bit set. Do you know
>> how to put that exact pattern of bits into a 4-byte integer in fortran?
>>
 

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