函数int（）向负无穷大（floor）舍入或为零？ [英] Function int() rounding towards negative infinity (floor) or zero?

问题描述

我看到为什么在C ++中，-1 / 2评估为0，但在Python中为-1？表示Python中的整数除法向无穷大舍入，即 floor 应用于结果。

I see Why is -1/2 evaluated to 0 in C++, but -1 in Python? says integer division rounds towards infinity in Python, namely, floor is applied to the result.

我认为 int（value）也会做类似 floor ，而我在实践中得到 int（-1.5）== -1 ，预计 -2 在我的脑海里。

I thought int(value) would also do something like floor, while I get int(-1.5) == -1 in practice, which was expected to be -2 in my mind.

所以问题是：为什么规则在整数除法和函数 int（）之间不一致？有没有合理的解释？

So question is: why rules are inconsistent between integer division and function int()? Is there any reasonable explanation?

推荐答案

int（）删除小数零件;它没有做任何舍入。来自文档：

int() removes the decimal component; it doesn't do any rounding. From the documentation:

如果x是浮点数，则转换将截断为零。

If x is floating point, the conversion truncates towards zero.

用于转动 float 进入 int 这完全是逻辑行为。这是不分区，地板或其他。

For turning a float into an int this is entirely logical behaviour. This is not division, flooring or otherwise.

// 楼层划分运营商否则显然 地板，而不是截断。在Python 2中，对于两个整数操作数， / 除法也是底层。再次文档：

The // floor division operator otherwise clearly does floor, not truncate. In Python 2, for two integer operands, the / division also floors. The documentation again:

结果是数学除法的结果，其中'floor'函数应用于结果

the result is that of mathematical division with the ‘floor’ function applied to the result

其中 math.floor（） 记录为：

where math.floor() is documented as:

返回 x 的最低值一个浮点数，小于或等于 x 的最大整数值。

Return the floor of x as a float, the largest integer value less than or equal to x.

我认为这里没有矛盾;划分楼层，将浮点数转换为整数截断值。

I see no inconsistency here; division floors, turning floats to integers truncates.

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