如何附加到python 3中的字节
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How to append to bytes in python 3
提问by felixbade
I have some bytes.
我有一些bytes。
b'\x01\x02\x03'
And an intin range 0..255.
并且int在范围0..255。
5
Now I want to append the intto the byteslike this:
现在我想将 附加int到bytes这样的：
b'\x01\x02\x03\x05'
How to do it? There is no appendmethod in bytes. I don't even know how to make the integer become a single byte.
怎么做？中没有append方法bytes。我什至不知道如何使整数成为单个字节。
>>> bytes(5)
b'\x00\x00\x00\x00\x00'
采纳答案by simonzack
bytesis immutable. Use bytearray.
bytes是不可变的。使用bytearray.
xs = bytearray(b'\x01\x02\x03')
xs.append(5)
回答by Ashwini Chaudhary
First of all passing an integer(say n) to bytes()simply returns an bytes string of nlength with null bytes. So, that's not what you want here:
首先传递一个整数（比如n）来bytes()简单地返回一个n长度为空字节的字节字符串。所以，这不是你想要的：
Either you can do:
你可以这样做：
>>> bytes([5]) #This will work only for range 0-256.
b'\x05'
Or:
或者：
>>> bytes(chr(5), 'ascii')
b'\x05'
As @simonzack already mentioned that bytes are immutable, so to update(or better say re-assign it to a new string) its value you need to use the +=operator.
正如@simonzack 已经提到字节是不可变的，因此要更新（或者更好地将其重新分配给新字符串）它的值，您需要使用+=运算符。
>>> s = b'\x01\x02\x03'
>>> s += bytes([5]) #or s = s + bytes([5])
>>> s
b'\x01\x02\x03\x05'
>>> s = b'\x01\x02\x03'
>>> s += bytes(chr(5), 'ascii') ##or s = s + bytes(chr(5), 'ascii')
>>> s
b'\x01\x02\x03\x05'
Help on bytes():
帮助bytes()：
>>> print(bytes.__doc__)
bytes(iterable_of_ints) -> bytes
bytes(string, encoding[, errors]) -> bytes
bytes(bytes_or_buffer) -> immutable copy of bytes_or_buffer
bytes(int) -> bytes object of size given by the parameter initialized with null bytes
bytes() -> empty bytes object
Construct an immutable array of bytes from:
- an iterable yielding integers in range(256)
- a text string encoded using the specified encoding
- any object implementing the buffer API.
- an integer
Or go for the mutable bytearrayif you need a mutable object and you're only concerned with the integers in range 0-256.
或者，bytearray如果您需要一个可变对象并且您只关心 0-256 范围内的整数，则选择可变对象。
