如何将浮点数转换为十六进制 [英] How to convert a float into hex

问题描述

在Python中，我需要将一堆浮点数转换为十六进制.必须填充为零(例如0x00000010而不是0x10).就像 http://gregstoll.dyndns.org/~gregstoll/floattohex/一样. (遗憾的是，我无法在平台上使用外部库，因此无法使用该网站上提供的库)

In Python I need to convert a bunch of floats into hexadecimal. It needs to be zero padded (for instance, 0x00000010 instead of 0x10). Just like http://gregstoll.dyndns.org/~gregstoll/floattohex/ does. (sadly i can't use external libs on my platform so i can't use the one provided on that website)

最有效的方法是什么?

推荐答案

这在python中有点棘手，因为它不希望将浮点 value 转换为(十六进制)整数.相反，您尝试解释 IEEE 754 的二进制表示形式浮点值为十六进制.

This is a bit tricky in python, because aren't looking to convert the floating-point value to a (hex) integer. Instead, you're trying to interpret the IEEE 754 binary representation of the floating-point value as hex.

我们将使用内置的 struct 库.

We'll use the pack and unpack functions from the built-in struct library.

A float是32位.我们首先将其pack转换为二进制 ¹ 字符串，然后将unpack用作int.

A float is 32-bits. We'll first pack it into a binary¹ string, and then unpack it as an int.

def float_to_hex(f):
    return hex(struct.unpack('<I', struct.pack('<f', f))[0])

float_to_hex(17.5)    # Output: '0x418c0000'

我们可以对double做同样的事情，知道它是64位:

We can do the same for double, knowing that it is 64 bits:

def double_to_hex(f):
    return hex(struct.unpack('<Q', struct.pack('<d', f))[0])

double_to_hex(17.5)   # Output: '0x4031800000000000L'

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^{1-表示一串原始字节； 不是一串一和零的东西.}

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^{1 - Meaning a string of raw bytes; not a string of ones and zeroes.}

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