/*

书中的公式是w位的公式(mod 就是为了截断成w位),我们现在做的是2w位中的前w位

注意书上这句话:由于模运算符,所有带有权重和的项都丢掉

对应到本题

该项除以后还是超过了2w位所以被丢弃了,因为题目说了只有2w位

这个式子除以就是我们想要的最终结果

函数signed_high_prod 的结果((int64_t) x * y )>> 32

对应的是x*y/   

函数unsigned_high_prod的结果中x * sig_y + y * sig_x

对应的是 

*/

#include <stdio.h>
#include <stdint.h>

#define TMax 2147483647
#define TMin (-TMax - 1)

int signed_high_prod(int x, int y);
unsigned unsigned_high_prod(unsigned x, unsigned y);

int main(void)
{

	unsigned x = 4294967295;
	unsigned y = 4294967295;
	printf("x*y\t\t%x\n",unsigned_high_prod(x,y));

}

int signed_high_prod(int x, int y) 
{
	printf("x*ysigned\t%8llx\n",(int64_t) x * y );
	printf("x*ysigned>>\t%8llx\n",((int64_t) x * y )>> 32);
    return ((int64_t) x * y )>> 32;
}

unsigned unsigned_high_prod(unsigned x, unsigned y) {

  unsigned signed_prod = signed_high_prod(x, y);

  int sig_x = x >> 31;//右移时x还是无符号所以能得到掩码
  int sig_y = y >> 31;
  return signed_prod + x * sig_y + y * sig_x;
}