>Shankar Unni <[EMAIL PROTECTED]> writes:
>
> > Well, what is it supposed to do? Technically, according to the C standard,
> > anyway, this operation has undefined effects, because you are converting a
> > negative number ((signed long) (y - z) * 10) into an unsigned number in
> > order to add to "x".
>
>Not quite. The conversion from unsigned to signed int *is* well-defined.
For some implementations.
>Chapter and verse: ANSI, 3.2.1.2, lines 23-27:
>
> "When a signed integer is converted to an unsigned integer with
> equal or greater size, if the value of the signed integer is
> nonnegative, its value is unchanged. Otherwise: if the unsigned
> integer has greater size, the signed integer is first promoted to
> the signed integer corresponding to the unsigned integer; the
> value is converted to unsigned by adding to it one greater than
> the largest number that can be represented in the unsigned integer
> type. [footnote 28]"
And if the unsigned integer does not have greater size?
> [footnote 28] In a two's-complement representation, there is no
> actual change in the bit pattern except filling the high-order
> bits with copies of the sign bit if the unsigned integer has
> greater size."
>
>So an expression like "(unsigned int) 1 + -10" is officially
>well-defined and has the value UINT_MAX - 8.
Assuming two's-complement representation.
>The irony is that an expression such as "(unsigned int) 10 + -10"
>trips over integer overflow and is hence officially undefined, though
>most implementations don't notice integer overflow and will "do what
>you expect" by producing a result of 0.
Yes, I agree.
The irony is quite thick.
-- Marshall
"The era of big government is over."
Bill Clinton, State of the Union Address, January 23, 1996
Marshall Clow Adobe Systems <mailto:[EMAIL PROTECTED]>
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