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Closed as not planned
Description
Bug report
Bug description:
def f1(s=''):
print(f'f1: {s}')
s += 'a'
return s
def f2(l=[]):
print(f'f2: {l}')
l.append('a')
return l
def f3(s=set()):
print(f'f3: {s}')
s.add('a')
return s
for i in range(4):
print(f'1a: {i}')
print(f'1b: {f1()}')
for i in range(4):
print(f'2a: {i}')
print(f'2b: {f2()}')
for i in range(4):
print(f'3a: {i}')
print(f'3b: {f3()}')Executing this code in python 3.13.0 will show you this output:
1a: 0
f1:
1b: a
1a: 1
f1:
1b: a
1a: 2
f1:
1b: a
1a: 3
f1:
1b: a
2a: 0
f2: []
2b: ['a']
2a: 1
f2: ['a']
2b: ['a', 'a']
2a: 2
f2: ['a', 'a']
2b: ['a', 'a', 'a']
2a: 3
f2: ['a', 'a', 'a']
2b: ['a', 'a', 'a', 'a']
3a: 0
f3: set()
3b: {'a'}
3a: 1
f3: {'a'}
3b: {'a'}
3a: 2
f3: {'a'}
3b: {'a'}
3a: 3
f3: {'a'}
3b: {'a'}
As you can see, a new string is created for each call of f1 but no new list for f2 or no new set for f3.
I was expecting something like:
1a: 0
f1:
1b: a
1a: 1
f1:
1b: a
1a: 2
f1:
1b: a
1a: 3
f1:
1b: a
2a: 0
f2: []
2b: ['a']
2a: 1
f2: []
2b: ['a']
2a: 2
f2: []
2b: ['a']
2a: 3
f2: []
2b: ['a']
3a: 0
f3: set()
3b: {'a'}
3a: 1
f3: set()
3b: {'a'}
3a: 2
f3: set()
3b: {'a'}
3a: 3
f3: set()
3b: {'a'}
CPython versions tested on:
3.13
Operating systems tested on:
Linux
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