c++ - 安全地舍入到下一个更小的倍数

Question

考虑:

int a, b;
int c = b * (a / b)

我认为它对正数的作用应该很清楚:c 被设置为 b 的倍数，b 相对于 a 的下一个较小值。

例子:令b = 2，则:

a = 0 => c = 0
a = 1 => c = 0
a = 2 => c = 2
a = 3 => c = 2

然而，对于负数，它以相反的方式工作:它取 b 的下一个更大(负数较小)的倍数。 同时选择下一个更小(更负)的 b 倍数的最佳方法是什么？

Answer 1

怎么样:

#include <cstdlib>
#include <iostream>

int round_down(int val, int unit) {
  std::div_t div_result = std::div(val, unit);
  int candidate = div_result.quot * unit;
  return candidate <= val ? candidate : 
         unit >= 0 ? candidate - unit :
         candidate + unit;
}

int main() {
  std::cout << round_down(9, 7) << std::endl;
  std::cout << round_down(13, 2) << std::endl;
  std::cout << round_down(15, 3) << std::endl;

  std::cout << round_down(-9, 7) << std::endl;
  std::cout << round_down(-13, 2) << std::endl;
  std::cout << round_down(-15, 3) << std::endl;

  std::cout << round_down(9, -7) << std::endl;
  std::cout << round_down(13, -2) << std::endl;
  std::cout << round_down(15, -3) << std::endl;

  std::cout << round_down(-9, -7) << std::endl;
  std::cout << round_down(-13, -2) << std::endl;
  std::cout << round_down(-15, -3) << std::endl;
}

请注意，在 C++11 之前，内置除法运算符 / 的舍入方向是实现定义的。所以，我改用 std::div()。引自 cppreference (强调我的):

The binary operator / divides the first operand by the second (after usual arithmetic conversions).
For integral operands, it yields the algebraic quotient.
The quotient is rounded in implementation-defined direction. (until C++11)
The quotient is truncated towards zero (fractional part is discarded). (since C++11)
...
Note: Until C++11, if one or both operands to binary operator % were negative, the sign of the remainder was implementation-defined, as it depends on the rounding direction of integer division. The function std::div provided well-defined behavior in that case.