我有一本需要其他功能參考的字典。我沒有將其設為全域，而是將字典作為引數傳遞。
def functionA(dictionary):
dictionary = {}
functionB(dictionary)
print(f"In A, after B, length: {len(dictionary)}")
def functionB(dictionary):
dictionary["A"] = "some value"
dictionary["B"] = "another value"
def main():
dictionary = {}
functionA(dictionary)
print(f"Back in main. Length: {len(dictionary)}")
if __name__ == "__main__":
""" This is executed when run from the command line """
main()
在上面的代碼中，main()初始化dictionary為空并將其傳遞給functionA()，這也（冗余地）將其設定為空并將其傳遞給functionB()，從而添加了幾個條目。
從main()回傳后functionA()，我預計dictionary將添加兩個條目functionB()。然而，它是空的。輸出是：
In A, after B, length: 2
Back in main. Length: 0
但是，如果我洗掉了多余dictionary = {}的functionA部分（即上述代碼塊的第二行），問題就消失了，我得到以下輸出：
In A, after B, length: 2
Back in main. Length: 2
我不明白為什么中的dictionary = {}陳述句functionA()會導致隨后添加的專案沒有被轉移回main(). 有什么見解嗎？我覺得這里有一些基本的東西我一無所知。
uj5u.com熱心網友回復：
關鍵是dictionary = {}不要清空原始字典。相反，它會創建一個新字典。如果我們將原始字典系結到不同的名稱，則更容易看出這一點：
def functionA(dictionary):
original = dictionary
dictionary = {}
functionB(dictionary)
print(f"In A, after B, length: {len(dictionary)}")
print(f"Original argument: {len(original)}")
In A, after B, length: 2
Original argument: 0
Back in main. Length: 0
創建main并傳遞給的原始字典functionA永遠不會被修改，因此它仍然是空的。傳遞給的dictionary內部是一個全新的物件（與 分開），它無法訪問，因為它從未被編輯過。functionAfunctionBoriginalmainreturn
如果要通過傳遞對原始字典的參考來修改原始字典，則需要確保使用 mutator 方法而不是重新系結名稱，例如：
def functionA(dictionary):
dictionary.clear()
functionB(dictionary)
print(f"In A, after B, length: {len(dictionary)}")
In A, after B, length: 2
Back in main. Length: 2
您還可以通過始終return設定您希望呼叫者接收的任何資料來避免這種型別的混淆：
def functionA(dictionary):
dictionary = {}
dictionary = functionB(dictionary)
print(f"In A, after B, length: {len(dictionary)}")
return dictionary
def functionB(dictionary):
dictionary["A"] = "some value"
dictionary["B"] = "another value"
return dictionary
def main():
dictionary = {}
dictionary = functionA(dictionary)
print(f"Back in main. Length: {len(dictionary)}")
In A, after B, length: 2
Back in main. Length: 2
uj5u.com熱心網友回復：
functionAdictionary將名稱重新分配給與引數中參考的記憶體不同的新記憶體。
您的代碼在邏輯上等同于：
def functionA(dictionary):
dictionary_a = {}
functionB(dictionary_a)
print(f"In A, after B, length: {len(dictionary_a)}")
def functionB(dictionary):
dictionary["A"] = "some value"
dictionary["B"] = "another value"
def main():
dictionary = {}
functionA(dictionary)
print(f"Back in main. Length: {len(dictionary)}")
if __name__ == "__main__":
""" This is executed when run from the command line """
main()
這應該使問題更加明顯。
您可以將dictionaryinmain()和dictionaryinfunctionA視為對同一字典的兩個單獨參考。當dictionaryinfunctionA被重新分配時，它不會影響dictionaryin的值main()。根據經驗，重新分配只會影響您放在陳述句左側的名稱。
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