Want to build a binary header block

[Larry Bates]

Bob Greschke wrote:
> This is the idea
> 
>    Block = pack("240s", "")
>    Block[0:4] = pack(">H", W)
>    Block[4:8] = pack(">H", X)
>    Block[8:12] = pack(">B", Y)
>    Block[12:16] = pack(">H", Z))
> 
> but, of course, Block, a str, can't be sliced.  The real use of this is a 
> bit more complicated such that I can't just fill in all of the "fields" 
> within Block in one giant pack().  I need to build it on the fly.  For 
> example the pack(">H", X) may be completely skipped some times through the 
> function.  There are many more fields in Block than in this example and they 
> are all kinds of types not just H's and B's.  What I really want is a C 
> struct union. :)
> 
> How would I do this?
> 
> Thanks!
> 
> Bob
> 
> 
When I have something like this I normally write a class
for it and make its __str__ method return the packed output.

Example (not tested, but you should get the drift):

import struct

class blockStruct:
    def __init__(self):
        self.hstring=240*" "
        self.W=None
        self.X=None
        self.Y=None
        self.Z=None
        return

    def __str__(self):
        rtnvals=[]
        rtnvals.append(struct.pack("240s", self.hstring)
        rtnvals.append(struct.pack(">H", W))
        .
        .
        .
        return ''.join(rtnvals)


Then in your main program

bS=blockStruct()
bs.hstring='this is a test'.ljust(240, ' ')
bs.W=12
bs.X=17
bs.Y=1
bs.Z=0

print bS

Seemed to be a good way that made debugging and understanding
easy for me.

-Larry