python - 如何从结构中打印字符串？

Question

所以我有字符串john。我把它打包成一个结构。当我打开它时，如何打印 john？目前它只打印j。如果我将字符串更改为 Sammy 或其他长度不同的名称，情况是否相同？我有 2 个函数来打包和解包结构。这就是我不需要担心 first_name 的长度的原因。该函数可以帮我完成。

结构基本是

• user_id(在本例中为 1)
• first_name(一个人的名字。这个字符串可以有不同的长度。在本例中是 john)

我的代码

from struct import *

def make_struct(user_id, first_name):
    return pack("is", user_id, first_name)

def deconstruct_struct(structure):
    return unpack("is", structure)

packed = make_struct(1, "john")
unpacked = deconstruct_struct(packed)

print(unpacked[1])

当前输出为:

j

Answer 1

您需要将字符串的长度添加到格式字符串中:

packed = pack("i4s", 1, "john")
unpacked = unpack("i4s", packed)
print(unpacked[1])
>> john

如果你需要一个可变长度的字符串 -> packing and unpacking variable length array/string using the struct module in python

编辑:

您的解决方案可以这样扩展:

from struct import *

def make_struct(user_id, first_name):
    first_name_length = len(first_name)
    fmt = "ii{}s".format(first_name_length) #generate format string with length of first_name
    return pack(fmt, user_id, first_name_length, first_name) #add the length to the pack

def deconstruct_struct(structure):
    user_id, first_name_length = unpack("ii", structure[:8]) #extract only userid and length from the pack
    fmt = "ii{}s".format(first_name_length) #generate the format string like above
    #return unpack(fmt, structure) #this would return a (user_id, length of first name, first_name) tuple
    return (user_id, unpack(fmt, structure)[2]) #this way, we return only the (user_id, first_name) tuple

packed = make_struct(1, "john")
unpacked = deconstruct_struct(packed)