inspect.stack() and frame
in [Python]
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From: Félix-Antoine Fortin on 11 Mar 2010 18:05 Given this code : # Experience with frame import sys import inspect def foo(): stack = inspect.stack() print "foo frame : " + str(hex(id(sys._getframe()))) 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? Or why the call to "stack = inspect.stack()" change the address of the frame?
From: Alf P. Steinbach on 11 Mar 2010 18:22 * F�lix-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
From: Félix-Antoine Fortin on 11 Mar 2010 19:03 On Mar 11, 6:22 pm, "Alf P. Steinbach" <al...(a)start.no> wrote: > * F lix-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. > My bad. > > > > > > 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.. > Okay, I thought I got that when I read the id documentation, but now I get it. So the only to compare two ids, is by making sure their lifetimes overlap. In this case, instead of keeping the ids, I have to keep a reference on the frame to make sure it is still alive when I will compare it with a second one. Thanks! > > Or why the call to "stack = inspect.stack()" change the address of the > > frame? > > Does it? > Yeah it does... I always get N different id when I run foo() N times in a row. Actually, what you said about lifetime applies here too. Here is another quick snippet : import sys import inspect def foo(): stack = inspect.stack() return sys._getframe() def foo2(): stack = inspect.stack() del stack return sys._getframe() def bar(): inspect.stack() return sys._getframe() frame_foo = foo() frame_foo2 = foo2() frame_bar = bar() print sys.getrefcount(frame_foo) print sys.getrefcount(frame_foo2) print sys.getrefcount(frame_bar) Output : 3 2 2 So it seems that there is one more reference to the foo frame because only because of "stack = inspect.stack()", so its lifetime isn't done contrary to foo2 and bar frame, and the frame id of a foo frame is different for each call. Now, what is keeping a reference on foo frame? Thanks Alf, Felix
From: Gabriel Genellina on 12 Mar 2010 01:11 En Thu, 11 Mar 2010 21:03:07 -0300, F�lix-Antoine Fortin <felix.antoine.fortin(a)gmail.com> escribi�: > On Mar 11, 6:22 pm, "Alf P. Steinbach" <al...(a)start.no> wrote: >> * F lix-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()))) >> >> > 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. >> > > Okay, I thought I got that when I read the id documentation, but now I > get it. > So the only to compare two ids, is by making sure their lifetimes > overlap. In > this case, instead of keeping the ids, I have to keep a reference on > the frame > to make sure it is still alive when I will compare it with a second > one. The best way to compare object identities is using the 'is' operator: `a is b` returns true if and only if after evaluating both operands they are the very same object. id() may be misleading if you are not careful: py> id([]) == id([]) True py> [] is [] False >> > Or why the call to "stack = inspect.stack()" change the address of the >> > frame? >> >> Does it? > > Yeah it does... I always get N different id when I run foo() N times > in a row. Think again after reading the response below. > Actually, what you said about lifetime applies here too. Here is > another quick > snippet : > > import sys > import inspect > > def foo(): > stack = inspect.stack() > return sys._getframe() > > def foo2(): > stack = inspect.stack() > del stack > return sys._getframe() > > def bar(): > inspect.stack() > return sys._getframe() > > frame_foo = foo() > frame_foo2 = foo2() > frame_bar = bar() > > print sys.getrefcount(frame_foo) > print sys.getrefcount(frame_foo2) > print sys.getrefcount(frame_bar) > > Output : > 3 > 2 > 2 > > So it seems that there is one more reference to the foo frame because > only because of "stack = inspect.stack()", so its lifetime isn't done > contrary to foo2 and bar frame, and the frame id of a foo frame is > different for each call. > > Now, what is keeping a reference on foo frame? The foo frame keeps a reference to the 'stack' local variable (in its f_locals attribute), and 'stack' keeps a reference to the current frame too. This doesn't happen neither in foo2() nor bar(), where the local array is empty. inspect.stack() isn't special: any other reference to the current frame would have the same effect. Let's examine the simple example above: py> id([]) == id([]) True Python creates an empty list, takes its id, and discards the list. The list object is then ready to be re-used again (the interpreter keeps a list of free objects for many common types), so the right-hand side gets the very same list. The same thing happens with the frame object in your first examples foo2() and bar(): the frame object is discarded after leaving the function, and is ready to be used again in the next call. But foo() creates a circular reference - the frame object is still alive after leaving the first call, so the second call must use a new frame. (The garbage collector will, eventually, break the cycle and free those objects, but not very soon). -- Gabriel Genellina
From: Félix-Antoine Fortin on 12 Mar 2010 09:08
Thanks Gabriel, you resumed quite well what I did discovered after my second post by playing with the garbage collector module. > (The garbage collector will, > eventually, break the cycle and free those objects, but not very soon). I'm not very familiar with the Python garbage collector, so you may excuse my simple question, but how can it break the cycle? I guess the object will be freed at least when the program ends, but could it be before that? Is there a mechanisme in the garbage collector to detect circular references? Felix |