如何获得浮点值的小数部分? [英] How should I obtain the fractional part of a floating-point value?

问题描述

我有一个类型为 float 的变量 x ，我需要它的小数部分.我知道我可以用

I have a variable x of type float, and I need its fractional part. I know I can get it with

• x-floorf(x)或
• fmodf(x，1.0f)

我的问题:其中一个总是比另一个更可取吗?它们实际上是一样的吗?我可能会考虑第三种选择吗?

My questions: Is one of these always preferable to the other? Are they effectively the same? Is there a third alternative I might consider?

注意:

• 如果答案取决于我使用的处理器，则将其设置为x86_64，并且如果您可以详细说明其他会很好的处理器.
• 请确保并参考 x 的负值上的行为.我不介意这种行为，但我需要知道这种行为是什么.

• If the answer depends on the processor I'm using, let's make it x86_64, and if you can elaborate about other processors that would be nice.
• Please make sure and refer to the behavior on negative values of x. I don't mind this behavior or that, but I need to know what the behavior is.

推荐答案

我可能会考虑第三种选择吗?

Is there a third alternative I might consider?

有专用功能. modff 的存在是为了将数字分解为整数和小数部分./p>

There's the dedicated function for it. modff exists to decompose a number into its integral and fractional parts.

float modff(float arg，float * iptr);

将给定的浮点值 arg 分解为整数和分数零件，每个零件的类型和符号与 arg 相同.不可或缺的部分(以浮点格式)存储在 iptr 指向的对象中.

Decomposes given floating point value arg into integral and fractional parts, each having the same type and sign as arg. The integral part (in floating-point format) is stored in the object pointed to by iptr.

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