Re: Printf question.

Keith Thompson wrote:
Jack Klein <jackklein@xxxxxxxxxxx> writes: 
On 30 Dec 2005 11:21:25 -0800, "praeiji" <spacecaps@xxxxxxxxx> wrote
in comp.lang.c:
[...] 
unsigned char a = 200;
char b = 200;
printf( "%d %d", a, b );

gives :

200, -56

How comes? I didn't tell printf that the first argument was unsigned
and it detected it on its own. It doesn't seem possible with varargs.
How is it possible?

[snip]

In the case of integer types of lesser rank than int, the "integer
promotions" occur.  Since a signed int can hold the value 200, the
value in the unsigned char 'a' is converted to the int value 200.


Correction: since a signed int can hold any value of type unsigned
char, the value in the unsigned char 'a' is promoted to type int.


<annoying pedant>
What, even on a system with CHAR_BIT==16 and sizeof(int)==1 ?
</annoying pedant>

Of course, you are only likely to find such systems in the embedded world where you may well not have printf.

> The 
promoted type doesn't depend on the value that happens to be in the
variable; it depends on which type can hold *any* possible value of
the type.

You are, of course, correct that is is purely a matter of type, not the value the variable happens to contain.
--
Flash Gordon
Living in interesting times.
Although my email address says spam, it is real and I read it.
.

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