Can I access a list while it is being sorted in the list.sort()
b = ['b', 'e', 'f', 'd', 'c', 'g', 'a']
f = 'check this'
def m(i):
print i, b, f
return None
b.sort(key=m)
print b
this returns
b [] check this
e [] check this
f [] check this
d [] check this
c [] check this
g [] check this
a [] check this
Note that individual items of list b
is sent to function m
. But at m
the list b
is empty, however it can see the variable f
, which has same scope as list b
. Why does function m
print b
as []
?
Looking at the source code (of CPython, maybe different behaviour for other implementations) the strange output of your script becomes obvious:
/* The list is temporarily made empty, so that mutations performed
* by comparison functions can't affect the slice of memory we're
* sorting (allowing mutations during sorting is a core-dump
* factory, since ob_item may change).
*/
saved_ob_size = Py_SIZE(self);
saved_ob_item = self->ob_item;
saved_allocated = self->allocated;
Py_SET_SIZE(self, 0);
The comment says it all: When you begin sorting, the list is emptied. Well, it is "empty" in the eye of an external observer.
I quite like the term "core-dump factory".
Compare also:
b = ['b','e','f','d','c','g','a']
f = 'check this'
def m(i):
print i, b, f
return None
b = sorted(b, key= m)
print b