>>>>> John O'Hagan <resea...@johnohagan.com> (JO) wrote:
>JO> I guess what I meant was that if I type:
>JO> brian = Brian()
>JO> in the python shell and then hit return, it seems to me that
>JO> _somewhere_ (in the interpreter? I have no idea how it's done) it
>JO> must be written that the new Brian object will later be assigned
>JO> the name "brian", even as the process of creating the instance
>JO> begins. As you've just demonstrated, the actual assignment occurs
>JO> afterwards.
No, that's not how it works. There is no `preparation' for the
assignment and there is no reason for it. Look at the generated
byte code for this Python code:
def test():
class Brian:
def __init__(self):
print "I'm Brian!"
brian = Brian()
2 0 LOAD_CONST 1 ('Brian')
3 LOAD_CONST 3 (())
6 LOAD_CONST 2 (<code object Brian at 0x35728, file
"<stdin>", line 2>)
9 MAKE_FUNCTION 0
12 CALL_FUNCTION 0
15 BUILD_CLASS
16 STORE_FAST 0 (Brian)
5 19 LOAD_FAST 0 (Brian)
22 CALL_FUNCTION 0
25 STORE_FAST 1 (brian)
28 LOAD_CONST 0 (None)
31 RETURN_VALUE
You can see that the assignment (STORE_FAST) is done after the method
call (CALL_FUNCTION) and there is no reference to the variable name
before the call.
If you do a parallel assignment the relationship becomes even less:
brian, test = Brian(), 1
5 19 LOAD_FAST 0 (Brian)
22 CALL_FUNCTION 0
25 LOAD_CONST 3 (1)
28 ROT_TWO
29 STORE_FAST 1 (brian)
32 STORE_FAST 2 (test)
--
Piet van Oostrum <p...@cs.uu.nl>
URL: http://pietvanoostrum.com [PGP 8DAE142BE17999C4]
Private email: p...@vanoostrum.org
--
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