当a和b都小于c但a * b溢出时，如何计算a * b/c?

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假设uint是我的定点平台上最大的整数类型，我有:

Assuming that uint is the largest integral type on my fixed-point platform, I have:

uint func(uint a, uint b, uint c);

需要返回近似值a * b / c的

c的值大于a的值和b的值.

The value of c is greater than both the value of a and the value of b.

因此我们可以肯定地知道a * b / c的值将适合uint.

So we know for sure that the value of a * b / c would fit in a uint.

但是，a * b的值本身溢出了uint的大小.

However, the value of a * b itself overflows the size of a uint.

所以计算a * b / c值的一种方法是:

So one way to compute the value of a * b / c would be:

return a / c * b;

甚至:

if (a > b) return a / c * b; return b / c * a;

但是，c的值大于a的值和b的值.

However, the value of c is greater than both the value of a and the value of b.

所以上面的建议只会返回零.

So the suggestion above would simply return zero.

我需要按比例减小a * b和c，但是同样-问题是a * b溢出.

I need to reduce a * b and c proportionally, but again - the problem is that a * b overflows.

理想情况下，我将能够:

Ideally, I would be able to:

• 将a * b替换为uint(-1)
• 用uint(-1) / a / b * c替换c.

• Replace a * b with uint(-1)
• Replace c with uint(-1) / a / b * c.

但是无论我如何对表达式uint(-1) / a / b * c进行排序，我都会遇到问题:

But no matter how I order the expression uint(-1) / a / b * c, I encounter a problem:

由于uint(-1) / a / b ，
• uint(-1) / a / b * c被截断为零
• uint(-1) / a * c / b由于uint(-1) / a * c
而溢出
• uint(-1) * c / a / b由于uint(-1) * c
而溢出

• uint(-1) / a / b * c is truncated to zero because of uint(-1) / a / b
• uint(-1) / a * c / b overflows because of uint(-1) / a * c
• uint(-1) * c / a / b overflows because of uint(-1) * c

我如何应对这种情况以便找到a * b / c的近似值?

How can I tackle this scenario in order to find a good approximation of a * b / c?

当最大整数类型为uint64时，平台上没有诸如_umul128之类的东西.我最大的类型是uint，并且我不支持任何大于该类型的内容(既不在硬件级别，也不在某些预先存在的标准库中).

I do not have things such as _umul128 on my platform, when the largest integral type is uint64. My largest type is uint, and I have no support for anything larger than that (neither on the HW level, nor in some pre-existing standard library).

我最大的类型是uint.

针对众多重复的建议和评论:

In response to numerous duplicate suggestions and comments:

我没有一些较大的类型"可以用来解决这个问题.这就是为什么该问题的开头声明是:

I do not have some "larger type" at hand, which I can use for solving this problem. That is why the opening statement of the question is:

假设uint是我的定点平台上最大的整数类型

Assuming that uint is the largest integral type on my fixed-point platform

我假设在SW层(通过某些内置的标准库)或在HW层上都没有其他类型.

I am assuming that no other type exists, neither on the SW layer (via some built-in standard library) nor on the HW layer.

推荐答案

我已经建立了一种可以解决O(1)复杂性(无循环)的解决方案:

I've established a solution which work in O(1) complexity (no loops):

typedef unsigned long long uint; typedef struct { uint n; uint d; } fraction; uint func(uint a, uint b, uint c); fraction reducedRatio(uint n, uint d, uint max); fraction normalizedRatio(uint a, uint b, uint scale); fraction accurateRatio(uint a, uint b, uint scale); fraction toFraction(uint n, uint d); uint roundDiv(uint n, uint d); uint func(uint a, uint b, uint c) { uint hi = a > b ? a : b; uint lo = a < b ? a : b; fraction f = reducedRatio(hi, c, (uint)(-1) / lo); return f.n * lo / f.d; } fraction reducedRatio(uint n, uint d, uint max) { fraction f = toFraction(n, d); if (n > max || d > max) f = normalizedRatio(n, d, max); if (f.n != f.d) return f; return toFraction(1, 1); } fraction normalizedRatio(uint a, uint b, uint scale) { if (a <= b) return accurateRatio(a, b, scale); fraction f = accurateRatio(b, a, scale); return toFraction(f.d, f.n); } fraction accurateRatio(uint a, uint b, uint scale) { uint maxVal = (uint)(-1) / scale; if (a > maxVal) { uint c = a / (maxVal + 1) + 1; a /= c; // we can now safely compute `a * scale` b /= c; } if (a != b) { uint n = a * scale; uint d = a + b; // can overflow if (d >= a) // no overflow in `a + b` { uint x = roundDiv(n, d); // we can now safely compute `scale - x` uint y = scale - x; return toFraction(x, y); } if (n < b - (b - a) / 2) { return toFraction(0, scale); // `a * scale < (a + b) / 2 < MAXUINT256 < a + b` } return toFraction(1, scale - 1); // `(a + b) / 2 < a * scale < MAXUINT256 < a + b` } return toFraction(scale / 2, scale / 2); // allow reduction to `(1, 1)` in the calling function } fraction toFraction(uint n, uint d) { fraction f = {n, d}; return f; } uint roundDiv(uint n, uint d) { return n / d + n % d / (d - d / 2); }

这是我的考试:

#include <stdio.h> int main() { uint a = (uint)(-1) / 3; // 0x5555555555555555 uint b = (uint)(-1) / 2; // 0x7fffffffffffffff uint c = (uint)(-1) / 1; // 0xffffffffffffffff printf("0x%llx", func(a, b, c)); // 0x2aaaaaaaaaaaaaaa return 0; }