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Exercise 4

Without running this code, what will it print? Why?

dict1 = {
    'a': [1, 2, 3],
    'b': (4, 5, 6),
}

dict2 = dict(dict1)
dict1['a'][1] = 42
print(dict2['a'])

Solution

The code outputs:

[1, 42, 3]

As in the previous exercise, the constructor invocation dict(dict1) creates a new dict that contains the same key/value pairs as dict1. Thus, dict2 is not the same object as dict1.

On line 7, we reassign dict1['a'][1] to 42. Since dict1 and dict2 are different dicts, you might expect that mutating one of dict1's values would not impact dict2. However, that is not the case. The dicts are different objects but share value components since the dict constructor creates a shallow copy. Thus, mutations to dict1['a'] can be seen in dict2['a'].

This code demonstrates that two dicts with equal-value objects associated with every key may also share those objects. That isn't always the case, but you must understand what's happening in your code.

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