在两个大整数相乘期间捕获和计算溢出 [英] Catch and compute overflow during multiplication of two large integers

问题描述

我正在寻找一种有效(可选标准、优雅且易于实现)的解决方案来乘以相对较大的数字，并将结果存储到一个或多个整数中:

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 :

假设我有两个这样声明的 64 位整数:

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

uint64_t a = xxx, b = yyy; 

当我执行 a * b 时，如何检测操作是否导致溢出并在这种情况下将进位存储在某处?

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

请注意，我不想使用任何大型库，因为我对存储数字的方式有限制.

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

推荐答案

1.检测溢出:

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

固定除法 0(感谢 Mark！)

Fixed division by 0 (thanks Mark!)

<强>2.计算进位相当复杂.一种方法是将两个操作数拆分为半字，然后将 长乘法 应用于一半-单词:

2. 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 ((1ULL << 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;
}

为了确保部分和本身不会溢出，我们考虑最坏的情况:

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

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

令 B = 1 <<<32.然后我们有

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

我相信这会起作用 - 至少它可以处理 Sjlver 的测试用例.除此之外，它未经测试(甚至可能无法编译，因为我手头没有 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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