* Félix-Antoine Fortin:
Given this code :
# Experience with frame
import sys
import inspect
def foo():
stack = inspect.stack()
print "foo frame : " + str(hex(id(sys._getframe())))
hex returns a string. applying str is therefore redundant.
def foo2():
inspect.stack()
print "foo2 frame : " + str(hex(id(sys._getframe())))
def bar():
print "bar frame : " + str(hex(id(sys._getframe())))
foo()
foo()
foo2()
foo2()
bar()
bar()
Output example :
foo frame : 0x84d2c0
foo frame : 0x844bf0
foo2 frame : 0x898c90
foo2 frame : 0x898c90
bar frame : 0x898f70
bar frame : 0x898f70
Why are the ids (address) of the frame for each foo call not the same?
You're dealing with Python objects. You're not dealing with the computer's
machine stack. Whether you get the same id for two objects whose lifetimes don't
overlap depends on the implementation's memory and id allocation strategy.
Or why the call to "stack = inspect.stack()" change the address of the
frame?
Does it?
Cheers,
- Alf
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