Re: how to create unsigned hexadecimal initial value

["Leif Svalgaard" <leif@xxxxxxxx>]

From: <rjd@us.ibm.com>
Subject: Re: how to create unsigned hexadecimal initial value


> I think even the more recent examples are easy to explain in terms of the
> definition I gave for H literals.

Bob, here is your definition:
"
The H literal is used to specify a signed integer value.  The format is
H'h1h2h3h4', where h1, h2, h3, and h4 are hexadecimal digits.  If fewer
than 4 digits are specified, leading zeros (0x00) are prepended.  For
example,  H'FF' is equivalent to H'000000FF'.  The sign of the value is
determined by the value of h1.  The largest positive value that can
specified is 2147483647 (H'7FFFFFFF').  [Note:  this is not unlike integer
literals in  C++.  For example, 0xff in the following statement represents
the value 255, not -1:  int h = 0xff;]

The length of an H literal is the minimum number of bytes needed to
represent the value.  The length of a negative value is always 4 bytes.
"

I'll try to rewrite it the way I think you meant it:
"
The H literal is used to specify an integer value.  The format is
H'h1h2h3h4h5h6h7h8', where h1, h2, h3, h4, h5, h6, h7, and h8 are 
hexadecimal digits.  If fewer than 8 digits are specified, leading zeros 
(0x00) are prepended.  For example,  H'FF' is equivalent to H'000000FF'.  
The sign of the value is determined by the value of h1.  The largest 
positive value that can specified is 2147483647 (H'7FFFFFFF'). The
smallest negative value that can be specified is -2147483648
(H'80000000').
The length of an H literal is always 4 bytes.
".

The problem Gene had has really nothing to do with the H literal, but is
simply that one cannot initialize an unsigned integer with a negative
value regardless of its size.



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