我想乘两个32位数字a和b应该给出一个64位结果。与a和b的无符号32位整数,我提出了这样的:乘以两个32位数字......这个代码有什么问题
r = a * b
r = ((ah << 16) + al) * ((bh << 16) + bl)
= ((ah * 2^16) + al) * ((bh * 2^16) + bl)
= (ah * 2^16) * (bh * 2^16) + (ah * 2^16) * bl + al * (bh * 2^16) + al * bl
= (ah * bh * 2^32) + (ah * bl * 2^16) + (al * bh * 2^16) + (al * bl)
= ((ah * bh) << 32) + ((ah * bl) << 16) + ((al * bh) << 16) + (al * bl)
= ((ah * bh) << 32) + ((ah * bl + al * bh) << 16) + (al * bl)
我然后翻译到c如下
static void _mul64(unsigned int a, unsigned int b, unsigned int *hi, unsigned int *lo) {
unsigned int ah = (a >> 16), al = a & 0xffff,
bh = (b >> 16), bl = b & 0xffff,
rh = (ah * bh), rl = (al * bl),
rm1 = ah * bl, rm2 = al * bh,
rm1h = rm1 >> 16, rm2h = rm2 >> 16,
rm1l = rm1 & 0xffff, rm2l = rm2 & 0xffff,
rmh = rm1h + rm2h, rml = rm1l + rm2l;
rl = rl + (rml << 16);
rh = rh + rmh;
if(rml & 0xffff0000)
rh = rh + 1;
*lo = rl;
*hi = rh;
}
然而,当我运行该小测试其乘以= 0xFFFFFFFF与b = 0xFFFFFFFF并应产生0xFFFFFFFE00000001,我得到0xFFFFFFFD00000001而不是。我做错了吗?
int main(int argc, char **argv) {
unsigned int a, b, rl, rh;
unsigned long long r;
unsigned long long r1, r2, r3;
a = 0xffffffff;
b = 0xffffffff;
mul64(a, b, &rh, &rl);
r1 = ((unsigned long long) rh << 32) + rl;
r2 = (unsigned long long) a * b;
_mul64(a, b, &rh, &rl);
r3 = ((unsigned long long) rh << 32) + rl;
printf("a = 0x%08x, b = 0x%08x\n", (unsigned) a, (unsigned) b);
printf("_mul64: 0x%16llx\n", (unsigned long long) r3);
printf("a * b = 0x%16llx\n", (unsigned long long) r2);
return 0;
}
您应该打印出所有中间值的值,以确定出错的地方。 – 2012-07-11 17:44:52
@OliCharlesworth说过,调试是掌握的重要技能。 – TheZ 2012-07-11 17:45:29
你有没有想过当'rl = rl +(rml << 16);会溢出?你是否在所有适当的地方检测/补偿进位? – 2012-07-11 17:51:02