Question

import struct
def print_binary(number_string):
    bin = ""   
    if "." in number_string:
        number_string = float(number_string)
        bin += bin(struct.unpack('!i',struct.pack('!f', number_string)))
    else:
        num = int(number_string)  
        bin += "{0:b}".format(num)
    print(bin)

我在将带小数的数字转换为二进制（我的if语句）时遇到问题

test_values = "0 1234 -1234 12.4 0.6 -12.4 -0.6 -1234567890.123456789".split()

当它达到12.4时出现错误

Answer 1

我使用了递归。但是这里的某些值可能存在递归限制。

这是我手工使用的方法。

Fractional Part = .6875 in base 10
.6875*2 equals 1.375  whole number is 1
.375*2 equals 0.75    whole number is 0
.75*2 equals 1.5      whole number is 1
.5*2 equals 1.0       whole number is 1
Hence .6875D = .1011 in base 2

然而，想想12.4的价值。它将永远存在！

.4*2 = .8
.8*2 = 1.6
.6*2 = .2
.2*2 = .4
... and so on

所以我们必须在递归计算中设置一个限制。我做了10.虽然这听起来不是很多，但看看它有多精确。

我输入的12.4返回1100.0110011001，等于12.3994140625

非常接近，随意玩这个数字，取其上限等等。

以下是代码：

def get_integer_part(pre_decimal_string):
    # special case of negative 0 being a prefix "-0.6"
    if pre_decimal_string == "-0":
        return "-0"
    else:
        num = int(pre_decimal_string)
        return "{0:b}".format(num)


def get_decimal_part(post_decimal_string, string_builder, recurse):
    recurse += 1
    post_decimal_value = float("." + post_decimal_string)

    if post_decimal_value == 0 or recurse > 10:
        return string_builder
    else:
        temp_mult_str = str(post_decimal_value * 2)
        temp_mult_split = temp_mult_str.split(".")
        string_builder += temp_mult_split[0]
        return get_decimal_part(temp_mult_split[1], string_builder, recurse)


def print_binary(number_string):
    # handle case of no preceding 0 ".3" or
    if number_string[0] == ".":
        number_string = "0" + number_string

    # handle case of no preceding 0 and is negative
    if number_string[0:2] == "-.":
        number_string = "-0" + number_string[1:]

    if "." in number_string:
        str_split = number_string.split(".")
        print(get_integer_part(str_split[0]) + "." + str(get_decimal_part(str_split[1], "", 0)))
    else:
        print(get_integer_part(number_string))

test_values = "0 1234 -1234 12.4 0.6 -12.4 -0.6 -1234567890.123456789".split()
print(test_values)
for each in test_values:
    print_binary(each)

# special cases
print_binary("-.7")
print_binary(".67")

这是最终输出：

0
10011010010
-10011010010
1100.0110011001
0.1001100110
-1100.0110011001
-0.1001100110
-1001001100101100000001011010010.0001111110
-0.1011001100
0.1010101110

虽然我会以不同的方式做到这一点，但我尽可能地保留了您的代码和格式，以便您最容易学习。

将浮点数转换为二进制数

1 个答案: