Mixing
Multiple assignments in python
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up vote
8
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favorite
I need a clear explanation here.
Why does the following code work ?
foo1 = foo1[0] = [0]
Ok, I know assignments are done left to right.
How does python understand foo1 is a list?
Btw I know foo1 ends up as [[...]] its first element being itself.
python
edited 36 mins ago
timgeb
37k104875
asked 59 mins ago
Jeremie
643
add a comment |Â
up vote
8
down vote
favorite
I need a clear explanation here.
Why does the following code work ?
foo1 = foo1[0] = [0]
Ok, I know assignments are done left to right.
How does python understand foo1 is a list?
Btw I know foo1 ends up as [[...]] its first element being itself.
python
edited 36 mins ago
timgeb
37k104875
asked 59 mins ago
Jeremie
643
2
Depends on whatfoo1was originally. If looks like it was a list with at least one element in it.
– Makoto
55 mins ago
2
Is the question purely theoretical, or you found any use to this kind of construct? @Makoto it does not depend on the previous use / definition offoo1.
– norok2
52 mins ago
@norok2: Yeah it does. You can't referencefoo1[0]iffoo1isn't indexable (e.g. like an object which doesn't have that property overridden or a number).
– Makoto
47 mins ago
Theoretical question.
– Jeremie
46 mins ago
1
@Makoto Please copy paste OP's code into a fresh interpreter session.
– timgeb
29 mins ago
add a comment |Â
up vote
8
down vote
favorite
up vote
8
down vote
favorite
I need a clear explanation here.
Why does the following code work ?
foo1 = foo1[0] = [0]
Ok, I know assignments are done left to right.
How does python understand foo1 is a list?
Btw I know foo1 ends up as [[...]] its first element being itself.
python
edited 36 mins ago
timgeb
37k104875
asked 59 mins ago
Jeremie
643
I need a clear explanation here.
Why does the following code work ?
foo1 = foo1[0] = [0]
Ok, I know assignments are done left to right.
How does python understand foo1 is a list?
Btw I know foo1 ends up as [[...]] its first element being itself.
python
python
edited 36 mins ago
timgeb
37k104875
asked 59 mins ago
Jeremie
643
edited 36 mins ago
timgeb
37k104875
asked 59 mins ago
Jeremie
643
edited 36 mins ago
timgeb
37k104875
edited 36 mins ago
timgeb
37k104875
edited 36 mins ago
timgeb
37k104875
37k104875
asked 59 mins ago
Jeremie
643
asked 59 mins ago
Jeremie
643
asked 59 mins ago
Jeremie
643
643
2
Depends on whatfoo1was originally. If looks like it was a list with at least one element in it.
– Makoto
55 mins ago
2
Is the question purely theoretical, or you found any use to this kind of construct? @Makoto it does not depend on the previous use / definition offoo1.
– norok2
52 mins ago
@norok2: Yeah it does. You can't referencefoo1[0]iffoo1isn't indexable (e.g. like an object which doesn't have that property overridden or a number).
– Makoto
47 mins ago
Theoretical question.
– Jeremie
46 mins ago
1
@Makoto Please copy paste OP's code into a fresh interpreter session.
– timgeb
29 mins ago
add a comment |Â
2
Depends on whatfoo1was originally. If looks like it was a list with at least one element in it.
– Makoto
55 mins ago
2
Is the question purely theoretical, or you found any use to this kind of construct? @Makoto it does not depend on the previous use / definition offoo1.
– norok2
52 mins ago
@norok2: Yeah it does. You can't referencefoo1[0]iffoo1isn't indexable (e.g. like an object which doesn't have that property overridden or a number).
– Makoto
47 mins ago
Theoretical question.
– Jeremie
46 mins ago
1
@Makoto Please copy paste OP's code into a fresh interpreter session.
– timgeb
29 mins ago
2
2
Depends on what
foo1 was originally. If looks like it was a list with at least one element in it.– Makoto
55 mins ago
Depends on what
foo1 was originally. If looks like it was a list with at least one element in it.– Makoto
55 mins ago
2
2
Is the question purely theoretical, or you found any use to this kind of construct? @Makoto it does not depend on the previous use / definition of
foo1.– norok2
52 mins ago
Is the question purely theoretical, or you found any use to this kind of construct? @Makoto it does not depend on the previous use / definition of
foo1.– norok2
52 mins ago
@norok2: Yeah it does. You can't reference
foo1[0] if foo1 isn't indexable (e.g. like an object which doesn't have that property overridden or a number).– Makoto
47 mins ago
@norok2: Yeah it does. You can't reference
foo1[0] if foo1 isn't indexable (e.g. like an object which doesn't have that property overridden or a number).– Makoto
47 mins ago
Theoretical question.
– Jeremie
46 mins ago
Theoretical question.
– Jeremie
46 mins ago
1
1
@Makoto Please copy paste OP's code into a fresh interpreter session.
– timgeb
29 mins ago
@Makoto Please copy paste OP's code into a fresh interpreter session.
– timgeb
29 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
8
down vote
Because
foo1 = foo1[0] = [0]
is equivalent to
temp = [0]
foo1 = temp
foo1[0] = temp
it first evaluates expression and then assigns from left to right.
assigns the single resulting object to each of the target lists, from left to right
from here: https://docs.python.org/3/reference/simple_stmts.html#assignment-statements
Update: also see answers to similar question here: Multiple assignment and evaluation order in Python
answered 40 mins ago
Eduard
35816
add a comment |Â
up vote
0
down vote
Python variables know their types based on the type of variable assigned to it. It is a dynamically types language. In your code, the interpreter sees foo1 = foo1[0] = [0] and it fins a value at the end, which is [0]. It is a list with one element 0. Now, this gets assigned to the first element of the list foo1 through foo1[0] = [0]. But since foo1 is already declared, it creates an object which has a pointer to itself, and hence foo1 gets self-referenced infinitely, with the innermost list having 0.
The structure of the list foo1 will be the same when the code is foo1 = foo1[0].
The object foo1 has entered an infinite self-referenced loop.
answered 15 mins ago
a_ran
387
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
Because
foo1 = foo1[0] = [0]
is equivalent to
temp = [0]
foo1 = temp
foo1[0] = temp
it first evaluates expression and then assigns from left to right.
assigns the single resulting object to each of the target lists, from left to right
from here: https://docs.python.org/3/reference/simple_stmts.html#assignment-statements
Update: also see answers to similar question here: Multiple assignment and evaluation order in Python
answered 40 mins ago
Eduard
35816
add a comment |Â
up vote
8
down vote
Because
foo1 = foo1[0] = [0]
is equivalent to
temp = [0]
foo1 = temp
foo1[0] = temp
it first evaluates expression and then assigns from left to right.
assigns the single resulting object to each of the target lists, from left to right
from here: https://docs.python.org/3/reference/simple_stmts.html#assignment-statements
Update: also see answers to similar question here: Multiple assignment and evaluation order in Python
answered 40 mins ago
Eduard
35816
add a comment |Â
up vote
8
down vote
up vote
8
down vote
Because
foo1 = foo1[0] = [0]
is equivalent to
temp = [0]
foo1 = temp
foo1[0] = temp
it first evaluates expression and then assigns from left to right.
assigns the single resulting object to each of the target lists, from left to right
from here: https://docs.python.org/3/reference/simple_stmts.html#assignment-statements
Update: also see answers to similar question here: Multiple assignment and evaluation order in Python
answered 40 mins ago
Eduard
35816
Because
foo1 = foo1[0] = [0]
is equivalent to
temp = [0]
foo1 = temp
foo1[0] = temp
it first evaluates expression and then assigns from left to right.
assigns the single resulting object to each of the target lists, from left to right
from here: https://docs.python.org/3/reference/simple_stmts.html#assignment-statements
Update: also see answers to similar question here: Multiple assignment and evaluation order in Python
answered 40 mins ago
Eduard
35816
edited 35 mins ago
answered 40 mins ago
Eduard
35816
answered 40 mins ago
Eduard
35816
answered 40 mins ago
Eduard
35816
35816
add a comment |Â
add a comment |Â
up vote
0
down vote
Python variables know their types based on the type of variable assigned to it. It is a dynamically types language. In your code, the interpreter sees foo1 = foo1[0] = [0] and it fins a value at the end, which is [0]. It is a list with one element 0. Now, this gets assigned to the first element of the list foo1 through foo1[0] = [0]. But since foo1 is already declared, it creates an object which has a pointer to itself, and hence foo1 gets self-referenced infinitely, with the innermost list having 0.
The structure of the list foo1 will be the same when the code is foo1 = foo1[0].
The object foo1 has entered an infinite self-referenced loop.
answered 15 mins ago
a_ran
387
add a comment |Â
up vote
0
down vote
Python variables know their types based on the type of variable assigned to it. It is a dynamically types language. In your code, the interpreter sees foo1 = foo1[0] = [0] and it fins a value at the end, which is [0]. It is a list with one element 0. Now, this gets assigned to the first element of the list foo1 through foo1[0] = [0]. But since foo1 is already declared, it creates an object which has a pointer to itself, and hence foo1 gets self-referenced infinitely, with the innermost list having 0.
The structure of the list foo1 will be the same when the code is foo1 = foo1[0].
The object foo1 has entered an infinite self-referenced loop.
answered 15 mins ago
a_ran
387
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Python variables know their types based on the type of variable assigned to it. It is a dynamically types language. In your code, the interpreter sees foo1 = foo1[0] = [0] and it fins a value at the end, which is [0]. It is a list with one element 0. Now, this gets assigned to the first element of the list foo1 through foo1[0] = [0]. But since foo1 is already declared, it creates an object which has a pointer to itself, and hence foo1 gets self-referenced infinitely, with the innermost list having 0.
The structure of the list foo1 will be the same when the code is foo1 = foo1[0].
The object foo1 has entered an infinite self-referenced loop.
answered 15 mins ago
a_ran
387
Python variables know their types based on the type of variable assigned to it. It is a dynamically types language. In your code, the interpreter sees foo1 = foo1[0] = [0] and it fins a value at the end, which is [0]. It is a list with one element 0. Now, this gets assigned to the first element of the list foo1 through foo1[0] = [0]. But since foo1 is already declared, it creates an object which has a pointer to itself, and hence foo1 gets self-referenced infinitely, with the innermost list having 0.
The structure of the list foo1 will be the same when the code is foo1 = foo1[0].
The object foo1 has entered an infinite self-referenced loop.
answered 15 mins ago
a_ran
387
answered 15 mins ago
a_ran
387
answered 15 mins ago
a_ran
387
answered 15 mins ago
a_ran
387
387
add a comment |Â
add a comment |Â
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2
Depends on what
foo1was originally. If looks like it was a list with at least one element in it.– Makoto
55 mins ago
2
Is the question purely theoretical, or you found any use to this kind of construct? @Makoto it does not depend on the previous use / definition of
foo1.– norok2
52 mins ago
@norok2: Yeah it does. You can't reference
foo1[0]iffoo1isn't indexable (e.g. like an object which doesn't have that property overridden or a number).– Makoto
47 mins ago
Theoretical question.
– Jeremie
46 mins ago
1
@Makoto Please copy paste OP's code into a fresh interpreter session.
– timgeb
29 mins ago