正参数的 C 余数/模运算符定义

Question

我在我应该修复的程序中发现了一个函数，该函数的 mod函数定义:

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顺便说一句，永远都是积极的......

嗯？ if(i<0) ？

参数是 that

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

而且只是事后的想法

...; 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 (到目前为止一切顺利)，而且-1 。嗯……真的吗？ (让我们忽略这个事实，所提出的实现并不能处理所有情况。)

我知道模运算在数学上是这样的。但后来有人告诉我 C %事实上“不实现模运算符，而是实现余数”。

那么，C 是如何定义 % 的？运算符(operator)？

在初稿中我只找到

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总是6 ？或者可以是 -1 ，也？

Answer 1

根据标准:

When integers are divided, the result of the / operator is the algebraic quotient with any fractional part discarded. If the quotient a/b is representable, the expression (a/b)*b + a%b shall equal a. [ISO/IEC 9899:2011: 6.5.5]

这意味着 a 的符号保留在模数中。

 17 %  3  ->  2
 17 % -3  ->  2
-17 %  3  -> -2
-17 % -3  -> -2

<小时/>

所以不，6%7 不能是 -1，因为提醒必须具有相同的股息符号。