Somebody originally wrote:
> # I'm a newbie to C and the Palm OS and have only recently started to write
> # an application with Codewarrior Lite for Mac. So far reading
> # through this mailing list has given me lots of tips and insights.
>
> #result = 10 * 3600; // Here I get -29536 instead of 36000 when
Summarizing three follow-ups:
> I am a 100% amateur, but I'd write
>
> result = (Long) 10 * 3600;
-----
> OR
>
> result = (UInt32) 10 * 3600;
-----
> You need to cast the result as type Long. 10 and 3600 are type
> integer, so the result is calculated as type integer. If you
> cast it,
>
> result = (Long) 10 * 3600;
>
> you should get the correct result.
----- [ end of summaries ]
Sorry, but these are all wrong. Unless you tell the compiler otherwise,
it interprets integral literals as type int. A literal is a plain old
number, like 10 or 3600. (The period on that last sentence is
grammatical, and is NOT a decimal point.) An integral literal is one
that does not have a decimal point.
In the world of Palm OS, an int is 16 bits, and so has a range of
-32768 through +32767 inclusive.
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;
Actually, you need only append the L to one of the literals, because
the compiler will promote the other literal to the larger size before
doing the multiplication.
All this, plus much much more about the C language, is in the book "C:
A Reference Manual" by Harbison and Steele. I have found it to be an
indispensable resource, and I highly recommend it to EVERY person doing
C programming.
--
Roger Chaplin
<[EMAIL PROTECTED]>
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