Question

我试图将两个32位数a和b相乘，这应该得到64位结果。 a和b是无符号32位整数，我想出了这个：

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)

然后我将其翻译成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，当我运行这个小的测试，它将a = 0xFFFFFFFF与b = 0xFFFFFFFF相乘并且应该产生0xFFFFFFFE00000001时，我得到的是0xFFFFFFFD00000001。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;
}

Answer 1

您在此处添加了16位数量

rm1l = rm1 & 0xffff,    rm2l = rm2 & 0xffff,
rmh  = rm1h + rm2h,     rml  = rm1l + rm2l;

并将rml向左移动16位到rl，

rl = rl + (rml << 16);

当两个16位量的总和变为17位数时，丢弃进位。

此外，后一个和可能超过32位范围，在这种情况下，您将丢失另一个进位。

Answer 2

在初始化器中完成所有算法后，调试将变得困难。将所有这些计算移出初始化程序，然后在禁用优化的情况下编译代码。在调试器中逐步执行它，并确保每个步骤都生成您希望生成的值。当您在遵循手动解决的算法的同时浏览代码时，应该很容易发现代码和算法偏离的任何位置。

乘以两个32位数字......这段代码出了什么问题

2 个答案: