This is kind of an old thread, but in Python 3.2+ now you can simply say:
number = 100
number.to_bytes(4, byteorder = 'big')
or byteorder = 'little' as per your needs. Documentation here.
This is kind of an old thread, but in Python 3.2+ now you can simply say:
number = 100
number.to_bytes(4, byteorder = 'big')
or byteorder = 'little' as per your needs. Documentation here.
Have a look at the struct module. Probably all you need is struct.pack("I", your_int) to pack the integer in a string, and then place this string in the message. The format string "I" denotes an unsigned 32-bit integer.
If you want to unpack such a string to a tuple of for integers, you can use struct.unpack("4b", s):
>>> struct.unpack("4b", struct.pack("I", 100))
(100, 0, 0, 0)
(The example is obviously on a little-endian machine.)
python - How to convert a string of bytes into an int? - Stack Overflow
python - Convert 4 bytes to integer using struct - Stack Overflow
python - Converting 4 bytes to an integer - Stack Overflow
How is a integer mapped to 4 bytes? How to convert 4 bytes into integer in python - Stack Overflow
How can I convert bytes to an integer in Python?
Each method has its own advantages and use cases so that you can choose the most suitable one based on your programming requirements.
How can I use the int.from_bytes() function to convert bytes to an integer?
This will convert the bytes object to an integer and assign it to the 'integer' variable.
How can I convert bytes to an integer using unpacking with the struct module?
The format string can be created using the '>' or '<' character to indicate big-endian or little-endian byte order, followed by a letter representing the desired byte size (e.g., 'i' for 4-byte integer).
Once you have the format string, you can use the struct.unpack() function to unpack the bytes object and convert it to an integer.
In Python 3.2 and later, use
>>> int.from_bytes(b'y\xcc\xa6\xbb', byteorder='big')
2043455163
or
>>> int.from_bytes(b'y\xcc\xa6\xbb', byteorder='little')
3148270713
according to the endianness of your byte-string.
This also works for bytestring-integers of arbitrary length, and for two's-complement signed integers by specifying signed=True. See the docs for from_bytes.
You can also use the struct module to do this:
>>> struct.unpack("<L", "y\xcc\xa6\xbb")[0]
3148270713L
Assuming you're on at least 3.2, there's a built in for this:
int.from_bytes(bytes,byteorder, *,signed=False)...
The argument
bytesmust either be a bytes-like object or an iterable producing bytes.The
byteorderargument determines the byte order used to represent the integer. Ifbyteorderis"big", the most significant byte is at the beginning of the byte array. Ifbyteorderis"little", the most significant byte is at the end of the byte array. To request the native byte order of the host system, usesys.byteorderas the byte order value.The
signedargument indicates whether two’s complement is used to represent the integer.
## Examples:
int.from_bytes(b'\x00\x01', "big") # 1
int.from_bytes(b'\x00\x01', "little") # 256
int.from_bytes(b'\x00\x10', byteorder='little') # 4096
int.from_bytes(b'\xfc\x00', byteorder='big', signed=True) #-1024
Lists of bytes are subscriptable (at least in Python 3.6). This way you can retrieve the decimal value of each byte individually.
>>> intlist = [64, 4, 26, 163, 255]
>>> bytelist = bytes(intlist) # b'@\x04\x1a\xa3\xff'
>>> for b in bytelist:
... print(b) # 64 4 26 163 255
>>> [b for b in bytelist] # [64, 4, 26, 163, 255]
>>> bytelist[2] # 26