What is the output ?✍️

@PythonPr Output is : [1,2,3] Because Y=X.copy() is created the copy of X and store in Variable Y X.append(4) is added 4 at last index of X but not in Y We print(Y) that's why Output is [1,2,3]

@PythonPr The copy() method for y becomes a shallow copy that is bound to the original list object (i.e., [1,2,3]). When 4 is appended to x, y is still bound to [1,2,3]. So print(y) gives [1,2,3]

@PythonPr My bad. I misread it as copy library. list.copy returns completely new list Hence any change in x will not reflect in y Y will be [1,2,3] we can use id(x) and id(y) to validate there addresses both will be different

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@PythonPr Output - [1,2,3]

@PythonPr since when you used x.copy() copy constructor function to assign to y, y is a new object and is not pointing x as it would have if you merely assigned x to y. print(y) will print [1,2,3] print(x) will print [1,2,3,4] ans: [1,2,3]

@PythonPr shallow copy: y=[1, 2, 3]

@PythonPr Y will be [1,2,3] as x.copy returns new list.

@PythonPr 1,2,3

@PythonPr [1, 2, 3]

@PythonPr [1, 2, 3]

@PythonPr [1, 2, 3]

@PythonPr 1,2,3,4

@PythonPr [1, 2, 3]

@PythonPr [1,2,3]

@PythonPr [1,2,3,4]

@PythonPr y=[1,2,3]

@PythonPr 123

@PythonPr [1,2,3,4]

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@PythonPr The output should be [1,2,3], Y should remain untouched as list.copy() creates a shallow copy and here y points to a completely different list address so Y remains unchanged.

@PythonPr [1, 2, 3] You made a copy, not a reference. So changes to x don’t affect y. Since copy() creates a new list, modifying x doesn’t affect y, so the output remains [1, 2, 3]

@PythonPr [1, 2, 3, 4].

@PythonPr [1, 2, 3]

@PythonPr 出力結果は[1,2,3] Xに配列として[1,2,3]を格納、 その後yにその時点でのxの内容をコピーした。 そのさらに後でxに4を追加。 ここでxは[1,2,3,4]になるが、 print(y)ではyに値が反映された時点の[1,2,3]が出力される。

@PythonPr 1234

@PythonPr Output: [ 1 , 2 , 3 ] ... .@PythonPr

@PythonPr 45

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