shawn bright wrote:
> oh, sorry, i meant how to get the 0x0A27 out of two bytes
> a = 0x27 and b = 0x8A
I don't see what the number 0x0A27 has to do with bytes 0x27 and 0x8A,
but I'll assume you meant
0x0A for b.
>
> actually, in my script, i am not using the hex values at all, i have
> these because they are examples in the documentation of a machine i am
> talking to. i am actually using ord(a) and ord(b) to get digital
> values of the numbers
>
> so, i guess a better question is how to get 2599 from the ord(a) and
> ord(b), how do i put the two bytes together to make one number?
Well, if you have the bytes originally, instead of ording them, hex them.
so... given these bytes
bytelist = [chr(0x27),chr(0x8A)]
hexlist = []
for item in bytelist:
hexlist.append(hex(item)[2:].zfill(2))
print int('0x' + ''.join(hexlist) , 16)
A less roundabout way to do this:
given the number
0x0A,
this is stored in the system the same as the number 10.
>>> 0x0A
10
As you can see.
So having the ord values is okay, because they're the same values as the
hex of the bytes, just represented in a different base.
Okay, now we'll assume that a byte is 8 bits.
Because of this, we know that the number
0xFFEE
is the same as
(0xFF << 8) + 0xEE
In other words, the higher byte is shifted over by 8 bits, and the other
value is not shifted.
This idea could easily be expanded to encompass byte strings of any length.
Oh, and in the case of your example values,
>>> (0x0A << 8) + 0x27
2599
Note that the + operator has higher (or the same?) precedence than the
binary shift left.
This means that the value 0x0A will be shifted by 8+0x27 times if the
parenthesis are missing.
Other equivalent operations:
>>> 0x0A * 256 + 0x27
2599
>>> 0x0A * 2**8 + 0x27
2599
But like Kent said, you probably should use struct or something like that.
Just in case you're interested, I'm sending this e-mail as well.
HTH,
-Luke
>>
>
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