在C＃整数运算中，a / b / c总是等于a /（b * c）吗？ [英] In C# integer arithmetic, does a/b/c always equal a/(b*c)?

问题描述

设a，b和c为非大正整数。 a / b / c是否始终与C＃整数运算相等？/（b * c）？对我来说，在C＃中它看起来像：

Let a, b and c be non-large positive integers. Does a/b/c always equal a/(b * c) with C# integer arithmetic? For me, in C# it looks like:

int a = 5126, b = 76, c = 14;
int x1 = a / b / c;
int x2 = a / (b * c);

所以我的问题是： x1 == x2 对于所有a，b和c？

So my question is: does x1 == x2 for all a, b and c?

推荐答案

让 \ 表示整数除法（两个 int s之间的C＃ / 运算符）并让 / 表示通常的数学师。然后，如果 x，y，z 是正整数，我们忽略溢出，

Let \ denote integer division (the C# / operator between two ints) and let / denote usual math division. Then, if x,y,z are positive integers and we are ignoring overflow,

(x \ y) \ z
    = floor(floor(x / y) / z)      [1]
    = floor((x / y) / z)           [2]
    = floor(x / (y * z))
    = x \ (y * z)

其中

a \ b = floor(a / b)

从行跳转[1] 到行 [2] 上面解释如下。假设你有两个整数 a 和 b 和一个小数 f 在 [0,1] 的范围内。很容易看到

The jump from line [1] to line [2] above is explained as follows. Suppose you have two integers a and b and a fractional number f in the range [0, 1). It is straightforward to see that

floor(a / b) = floor((a + f) / b)  [3]

如果在行 [1] 中，您确定 a = floor（x / y）， f =（x / y） - floor（x / y），和 b = z ，然后 [3] 暗示 [1] 和 [2] 相等。

If in line [1] you identify a = floor(x / y), f = (x / y) - floor(x / y), and b = z, then [3] implies that [1] and [2] are equal.

您可以将此证明推广为负整数（仍然忽略溢出），但我会把它留给读者来保持简单。

You can generalize this proof to negative integers (still ignoring overflow), but I'll leave that to the reader to keep the point simple.

关于溢出的问题 - 请参阅Eric Lippert的答案以获得一个很好的解释！他还在他的博客文章中采取了更严格的方法并回答了一些问题。应该看看你是否觉得我太手了。

On the issue of overflow - see Eric Lippert's answer for a good explanation! He also takes a much more rigorous approach in his blog post and answer, something you should look into if you feel I'm being too hand-wavy.

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