大量繁殖，如何捕捉溢出 [英] multiplication of large numbers, how to catch overflow

问题描述

我要寻找一个有效的（可选标准，优雅，易于实现的）解决方案，以繁殖比较大的数字，并将结果存储到一个或几个整数：

I am looking for an efficient (optionally standard, elegant and easy to implement) solution to multiply relatively large numbers,and store the result into one or several integers :

让说我有声明如下两个64位的整数：

Let say I have two 64 bits integers declared like this :

uint64_t a = xxx, b = yyy; 

当我这样做 A * B ，我怎么能检测如果操作导致溢出，在这种情况下，店里进的地方？

When I do a*b, how can I detect if the operation result in an overflow and in this case store the carry somewhere ?

请注意，我不希望因为我有我存储数字的方式限制使用任何大数量的库。

Please note that I don't want to use any large-number library since I have constraints on the way I store the numbers.

感谢

推荐答案

检测：

x = a * b;
if (a != 0 && x / a != b) {
    // overflow handling
}

编辑：固定除数为0（！感谢马克）

Fixed division by 0 (thanks Mark!)

计算进是相当复杂。一种方法是将两个操作数分成半字，然后应用长乘法以半字

Computing the carry is quite involved. One approach is to split both operands into half-words, then apply long multiplication to the half-words:

uint64_t hi(uint64_t x) {
    return x >> 32;
}

uint64_t lo(uint64_t x) {
    return ((1 << 32) - 1) & x;
}

void multiply(uint64_t a, uint64_t b) {
    // actually uint32_t would do, but the casting is annoying
    uint64_t s0, s1, s2, s3; 

    uint64_t x = lo(a) * lo(b);
    s0 = lo(x);

    x = hi(a) * lo(b) + hi(x);
    s1 = lo(x);
    s2 = hi(x);

    x = s1 + lo(a) * hi(b);
    s1 = lo(x);

    x = s2 + hi(a) * hi(b) + hi(x);
    s2 = lo(x);
    s3 = hi(x);

    uint64_t result = s1 << 32 | s0;
    uint64_t carry = s3 << 32 | s2;
}

要看到，没有一个部分和本身可以溢出，我们认为最坏的情况：

To see that none of the partial sums themselves can overflow, we consider the worst case:

        x = s2 + hi(a) * hi(b) + hi(x)

让B = 1＆LT;＆LT; 32.然后我们有

Let B = 1 << 32. We then have

            x <= (B - 1) + (B - 1)(B - 1) + (B - 1)
              <= B*B - 1
               < B*B

我相信这将工作 - 至少它处理Sjlver的测试用例。除此之外，它是未经测试（可能甚至不会编译，因为我没有手头C ++编译器了）。

I believe this will work - at least it handles Sjlver's test case. Aside from that, it is untested (and might not even compile, as I don't have a C++ compiler at hand anymore).

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