Conversions/casts one more time

[Chris Torek]

>> 	printf("\tc = %x, (char)uc = %x, (char)ucf() = %x\n",
>> 	  c, (char)uc, (char)ucf());

In article <820 at mtund.UUCP> adam at mtund.UUCP (Adam V. Reed) writes:
>Ugh. %x expects an int, so the result of feeding it a char is,
>*and ought to be*, UNDEFINED.

printf is also a function, so it is not possible to hand it a char.
`char' exists only as a data type (lvalue), not as an expression
type (rvalue), so `printf("%x", (char)c);' sends printf (int)(char)c,
not (char)c.

As to the original question, a cast is (supposed to be) equivalent
to an assignment to a temporary variable of the given type.  The
printf() call above is therefore (supposed to be) equivalent to

	char t1, t2;

	t1 = uc;
	t2 = ucf();
	printf("\tc = %x, (char)uc = %x, (char)ucf() = %x\n",
		c, t1, t2);

Whether this sign extends 0xaa is machine, and sometimes compiler,
dependent.
-- 
In-Real-Life: Chris Torek, Univ of MD Comp Sci Dept (+1 301 454 7690)
UUCP:	seismo!mimsy!chris	ARPA/CSNet:	chris at mimsy.umd.edu