A little confusion between `char' and `unsigned char'

[ryanli]

I'm currently working on something that need bit operations and I decided to use unsigned char to manipulate. However, I've got some strange results.
Let's have a look at the code

#include <stdio.h>

int main()
{
	char a;
	unsigned char b;
	printf("%d %d\n", (int) sizeof(a), (int) sizeof(b));
	a = b = '\x7f';
	printf("%.2X %.2X\n", a, b);
	printf("%d %d\n", a, b);
	a = b = '\x80';
	printf("%.2X %.2X\n", a, b); //be cautious!
	printf("%d %d\n", a, b);
}

The result is this(under GCC 4.3):

1 1
7F 7F
127 127
FFFFFF80 80
-128 128

Why is the char of '\x80' shown as a 4-bit type while the result of sizeof function is still 1? And yet why will values from \x80 through \xff behave like this while values from \x00 through \x7f behave `normal'?
Great thanks if anyone could explain this to me.

[kbw]

As you know, the difference between a signed and unsigned type is how it's interpretted. But you should also appreciate that in C, chars are passed as int, when passed by value to a function.

So, printf("%.2X %.2X\n", a, b); and printf("%d %d\n", a, b); sees two ints. Both chars are extended to ints with the sign preserved.

[KarlisRepsons]

And, as it can be seen, if the signed variable's assigned value steps over the half of max value, it is treated as negative.

[jsmith]

Yes, the key is that %x requires an unsigned int, not a char, and thus the char is being sign extended to an int then output as unsigned. So 0x80 signed is -128, sign extended
to 32-bits is 0xFFFFFF80 (sign bit is propagated).

[ryanli]

Thanks a lot!