/*
 * What happens when the shift value is bigger than the amount of bits in the
 * variable?
 *
 * Some machines will only use the lower log2 w bits of the shift amount when
 * shifting a 2-bit falue.
 *
 * For example, shifting a 8-bit value by 10, would make the machine shift the
 * value for only 2 (a k mod w).
 *
 * This is not a rule though. Let's test it.
 *
 * Looks like linux GCC on 64bit will always shift the exactly number, and crop
 * only the exactly number of bits for the data type representation.
 *
 * For signed integers, if the shift is done right in the printf, gcc will use a
 * convert it to a signed 32-bits for the conversion.
 *
 * Try change the x in printf on line 32, to use "a << 10" directly.
 */
#include <stdio.h>

int main(void)
{
	/* 10101001 */
	char x, a = 0xA9;
	unsigned char y, b = 0xA9;


	printf("Signed Original value: 0x%.2x\n", a);
	x = a << 10;
	printf("Signed left shift by 10: 0x%.2x\n\n", x);


	printf("Unsigned Original value: 0x%.2x\n", b);
	y = b << 10;
	printf("Unsigned left shift by 10: 0x%.2x\n", y);

	return 0;
}