[EMAIL PROTECTED] wrote:
> In a message dated 2/25/00 10:08:31 PM Eastern Standard Time,
> [EMAIL PROTECTED] writes:
>
> > So, when the compiler sees
> > (Long) 10 * 3600
> > it multiplies an int 10 by an int 3600, then casts the result to a
> > long. Unfortunately, the 10 * 3600 = 36000, which is too large for an
> > int. The result is a (silent) math overflow. All you're doing is
> > casting the wrong value to a long!
> >
> > If you want to override the compiler's default interpretation of
> > literals, you have to tell it. Append an 'L' to a literal to make the
> > compiler interpret it as long:
> > long result;
> > result = 10L * 3600L;
>
> I politely disagree.
>
> (Long) 10 * 3600 will typecast 10 into a 32-bit value before doing the
> computation. Hence the result will be the same 32-bit value that you get by
> using 10L.
>
> -Pete
And K&R disagrees with you, see page 187:
An expression preceded by the parenthesized name of a data type causes conversion of
the value of the expression to the named
typed. The construction is call a cast.
If you not sure what's going on use more parentheses, ie. ((long)10)*3600 will do
what you want, otherwise the expression is
evaluated BEFORE the cast. Remember, the Palm is a 16-bit worl, so you have to do some
extra typing if you want 32 bit arithmetic.
chuck
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