I have a 开发者_运维百科list of bytes as follows
pkt_bytes = [ 0x02,0x07, 0xff,0xff ,0x00,0x03]
in the position 0xff,0xff
I want to put a 16bit short integer
How do I do it
Regards
You can use the struct
module to pack values into appropriate formats:
>>> pkt_bytes = [0x02, 0x07, 0xff, 0xff, 0x00, 0x03]
>>> myint = 123
>>> pkt_bytes[3:5] = [ord(b) for b in struct.pack("H",myint)]
>>> pkt_bytes
[2, 7, 255, 123, 0, 3]
By default this will use the native byte order but you can override this using modifiers to format string. Since your variable is called pkt_bytes
I'm guessing you want network (big-endian) byte order which is signified by a !
:
>>> struct.pack("!H",5000)
'\x13\x88'
Try:
>>> pkt_bytes.insert(3, 0xaa)
>>> help(pkt_bytes.insert)
Help on built-in function insert:
insert(...)
L.insert(index, object) -- insert object before index
The code below will replace every occurrence of 0xff
with 0x04
until there are no more 0xff
left in the list.
pkt_bytes = [0x02, 0x07, 0xff, 0xff ,0x00, 0x03]
while True:
try:
idx = pkt_bytes.index(0xff)
pkt_bytes[idx] = 0x04
except ValueError:
break
>>> pkt_bytes = [ 0x02,0x07, 0xff,0xff ,0x00,0x03]
>>> pkt_bytes[2:4] = [pkt_bytes[2] << 8 | pkt_bytes[3]]
>>> pkt_bytes
[2, 7, 65535, 0, 3]
精彩评论