正参数的C余数/模运算符定义 [英] C remainder/modulo operator definition for positive arguments

问题描述

我在应该修复的程序中找到了一个定义了 mod 函数的函数:

I found a function in a program I am supposed to fix that has a mod function defined:

int mod(int a, int b)
{
  int i = a%b;
  if(i<0) i+=b;
  return i;
}

有人告诉我 a 和 b 永远都是积极的...

I was told that a and b will always be positive by the way...

嗯? if(i< 0)?

参数是那

取模运算的结果是一个等价类，并且可以选择该类的任何成员作为代表

the result of the modulo operation is an equivalence class, and any member of the class may be chosen as representative

这只是事后的想法

...;但是，通常的代表是最小的正残基，是属于该类别的最小的非负整数，即欧几里得除法的余数.但是，其他约定也是可能的.

...; however, the usual representative is the least positive residue, the smallest nonnegative integer that belongs to that class, i.e. the remainder of the Euclidean division. However, other conventions are possible.

这意味着 6％7 可以返回 6 (到目前为止非常好)，但也可以返回 -1 .嗯...真的吗?(让我们忽略这样一个事实，即所提供的实现不能处理所有情况.)

That means that 6 % 7 could return 6 (so far so good), but also -1. Hrm... really? (Lets ignore the fact that the presented implementation does not handle all cases.)

我知道在数学上模运算是这样的.但是后来有人告诉我，C ％实际上确实不是实现模运算符，而是实现余数".

I know that it is mathematically true that the modulo operation is like this. But then someone else told me that the C % does in fact "not implement the modulo operator but the remainder".

那么，C如何定义％运算符?

So, how does C define the % operator?

在C-Draft中我只能找到

In the C-Draft I only find

/运算符的结果是第一个操作数除以第二;％运算符的结果是余数.在这两个操作中，如果第二个操作数为零，其行为是不确定的.

The result of the / operator is the quotient from the division of the first operand by the second; the result of the % operator is the remainder. In both operations, if the value of the second operand is zero, the behavior is undefined.

这是否意味着 6％7 始终是 6 ?或者也可以是 -1 ?

Does this mean, that 6 % 7 is always 6? Or can it be -1, too?

推荐答案

永远:

• a ==(a/b)* b + a％b
• abs(a％b)<abs(b)
• 如果 a 和 b 为正，则 a％b 为正.

• a == (a/b)*b + a%b
• abs(a%b) < abs(b)
• if a and b are positive, a % b is positive.

从C99开始，

• a/b == trunc(a/b)
• a％b 是 0 或带有 a 的符号.

• a/b == trunc(a/b)
• a%b is either 0 or has the sign of a.

认为 6％7 可能是 -1 的原因可能是因为缺少了 a 和 b的结果肯定会得到肯定，并且缺少C99中的更改.

Thinking that 6 % 7 could be -1 is probably due to missing the fact that the result for a and b positive has always been guaranteed and missing the change in C99.

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