CODEWITHSUNDEEP
pythonparameters
Tuan Le PN
Tuan Le PN

Reputation: 384

python parameters are passed by reference?

Many articles on Internet say that python parameters are passed by reference. But from this snippet, the variable d is unchanged after the function test() is called. It is different from C/C++. Could someone please explain it? Thanks.

def test(_d : dict):
    _d = dict()
    _d.update({'D': 4})
    print("Inside the function",_d)
    return

d = {'A': 1,'B':2,'C':3}
test(d)
print("outside the function:", d) # expected: {'D:4}

Upvotes: 1

Views: 107

Answers (1)

dawg
dawg

Reputation: 104092

Nothing complicated here.

When you have _d = dict() you just created a new name _d (local to the function test) for a new dict and lost the local name _d referring to the global dict d that passed to that function as _d initially:

def test(_d : dict):
    print("Passed _d id:",id(_d))
    _d = dict()
    print("new _d id:",id(_d))
    _d.update({'D': 4})
    print("Inside the function",_d)
    return

d = {'A': 1,'B':2,'C':3}
print('Passed d id:',id(d))
test(d)
print("outside the function:", d) # expected: {'A':1, 'B':2, 'C':3, 'D:4}

Prints:

Passed d id: 4404101632
Passed _d id: 4404101632
new _d id: 4405135680
Inside the function {'D': 4}
outside the function: {'A': 1, 'B': 2, 'C': 3}

Now try:

def test(_d : dict):
    print("Passed _d id:",id(_d))
    # _d = dict()
    print("new _d id?:",id(_d))
    _d.update({'D': 4})
    print("Inside the function",_d)
    return

d = {'A': 1,'B':2,'C':3}
print('Passed d id:',id(d))
test(d)
print("outside the function:", d) # expected: {'A':1, 'B':2, 'C':3, 'D:4}

Prints:

Passed d id: 4320473600
Passed _d id: 4320473600
new _d id?: 4320473600
Inside the function {'A': 1, 'B': 2, 'C': 3, 'D': 4}
outside the function: {'A': 1, 'B': 2, 'C': 3, 'D': 4}

Note in the second case the id is the same at all three places printed so it is the same object; in the first case, the id changes so you have a different object after _d = dict() is called.

Note:

Since d is global and mutable, this works as expected as well:

def test():
    d.update({'D': 4})

d = {'A': 1,'B':2,'C':3}
test()
print(d) # {'A': 1, 'B': 2, 'C': 3, 'D': 4}

Upvotes: 1

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