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<div class="moz-cite-prefix">Just to menion:<br>
<br>
I am NOT trying to get anything changed. It works fine for me as
it is.<br>
<br>
I would however like to widen/correct my understanding of the
matter<br>
<br>
On 14/01/2013 16:03, Martin wrote:<br>
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<blockquote cite="mid:50F42C69.4040409@mfriebe.de" type="cite">
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not sure if I follow. If "or" performs on a "set of bits" (rather
than a number), and a set (not being a number) is neither signed
or unsigned, then before the "OR" both operands (independent of
being signed or not) will be cast to a set. The result is a set.
The set can be converted to either signed or unsigned. The set has
no range checking.<br>
<br>
result := bit_set32(a) or bit_set32(b)<br>
<br>
If result has also 32 (or more) bits, then no range check error
can occur. In the set, the high-bit has no meaning. It is neither
part of the (unsigned positive) numer, nor is it sign indicator.
This meaning is only applied by casting it to a numeric type.<br>
<br>
Same as<br>
var i: set of (b0, b1, b2 .... b31);<br>
signed := integer(i);<br>
unsigned := cardinal(i);<br>
<br>
<br>
<br>
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<pre wrap="">_______________________________________________
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</pre>
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