Round-Integer-Division 整數除法 數值簡化

There are different situations of interger division for C++ and Python. How to implement round up/down of interger division?

C++ 和 Python 在整數除法好像會出現一些不一樣的結果？來討論看看吧~

In Mathematics 數學上

Quotient & Remainder 商 與 餘

• 10 / 3 = 3 ... 1
• 10 / -3 = -3 ... 1
• -10 / 3 = -4 ... 2
• -10 / -3 = 4 ... 2

float 浮點數

• 10 / 3 = 3.3333
• 10 / -3 = -3.3333
• -10 / 3 = -3.3333
• -10 / -3 = 3.3333

In C++

• 10 / 3 = 3
• 10 / -3 = -3
• -10 / 3 = -3
• -10 / -3 = 3

(I think) C++ integer division will round towards zero (or say truncate, or round away from infinity)

Hence, we can infer the remainder (modulus) by the result above.

(我猜) C++ 的整數除法會是無條件捨去，所以可以由此而知，C++ 的模除 (modulus) 會如下。

• 10 % 3 = 1 // 10 = 3 * 3 + 1
• 10 % -3 = 1 // 10 = -3 * -3 + 1
• -10 % 3 = -1 // -10 = 3 * -3 - 1
• -10 % -3 = -1 // -10 = -3 * -3 - 1

In Python

• 10 / 3 = 3
• 10 / -3 = -4
• -10 / 3 = -4
• -10 / -3 = 3

Note: if use python3, 10/3 may get float 3.333. Hence, we should use 10**//**3 and get 3.0 where // means floor div

若使用 python3 ，會得到浮點數 (float) 的結果，所以要用 // 來得到下取整除法

(I think) Python integer division will round down (or say floor, or round towards negative infinity)

Hence, we can infer the remainder (modulus) by the result above.

(我猜) Python 的整數除法會是下取整，所以可以由此而知，Python 的模除 (modulus) 會如下。

• 10 % 3 = 1 // 10 = 3 * 3 + 1
• 10 % -3 = -2 // 10 = -3 * -4 - 2
• -10 % 3 = 2 // -10 = 3 * -4 + 2
• -10 % -3 = -1 // -10 = -3 * -3 - 1

How to Round Up by C++

If the result is positive, the integer division will round down.

If a/b is positive, we can use #define ROUND_UP_DIV(a,b) ((a + b - 1) / b).

However, overflow may happen and makes the answer wrong.

如果 a/b 是正數時，原本的整數除法會得到下取整，所以要使用 #define ROUND_UP_DIV(a,b) ((a + b - 1) / b) 來得到上取整

不過要注意有可能會溢位 (overflow)，而導致答案錯誤

How to Round Down by C++

If the result is negative, the integer division will round up.

1. #define ROUND_DOWN_DIV(a,b) ((a - b + 1) / b) if a/b is negative.

2. #define ROUND_DOWN_DIV(a,b) ((a - (((a % b) + b) % b)) / b) for all cases of a/b without "if" branches.

However, overflow may happen and makes the answer wrong.

3. #define ROUND_DOWN_DIV2_to_n(a,n) (a << n) if b = 2ⁿ

如果 a/b 是負數時，原本的整數除法會得到上取整，所以要使用 #define ROUND_DOWN_DIV(a,b) ((a - b + 1) / b) 來得到下取整

也可以使用 #define ROUND_DOWN_DIV(a,b) ((a - (((a % b) + b) % b)) / b) 以避免使用 if 來判斷正負

同樣要注意有可能會溢位 (overflow)，而導致答案錯誤

若 b = 2ⁿ, 最好的用法是 #define ROUND_DOWN_DIV2_to_n(a,n) (a << n)