{"id":24445,"date":"2022-12-03T01:43:00","date_gmt":"2022-12-02T20:13:00","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-whats-the-difference-between-class-reference-and-class-object-when-the-function-return-class-object-in-c\/"},"modified":"2022-12-03T01:43:00","modified_gmt":"2022-12-02T20:13:00","slug":"solved-whats-the-difference-between-class-reference-and-class-object-when-the-function-return-class-object-in-c","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-whats-the-difference-between-class-reference-and-class-object-when-the-function-return-class-object-in-c\/","title":{"rendered":"[Solved] What’s the difference between class reference and class object when the function return class object in C++?"},"content":{"rendered":"

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Abc &obj11 = fun1();\n<\/code><\/pre>\n
This line makes the program ill-formed;<\/strong> the C++ specification forbids temporaries to be bound to a reference that is not const<\/code>.  A compliant C++ compiler would emit an error to this effect.  Presumably you are using the Microsoft Visual C++ compiler, which is well-known for allowing the binding of a temporary to a non-const<\/code> reference in contradiction to the C++ specification.<\/p>\n
const Abc &obj11 = fun1();\n<\/code><\/pre>\n
This line would be well-formed, and it causes the temporary to be bound to the reference.<\/p>\n
In any case, when binding a temporary to a reference the temporary will have its lifetime extended to match the lifetime of the reference, so when the reference goes out of scope the temporary will also be destroyed.  In other words, in terms of lifetime semantics it’s equivalent to the line Abc obj1 = fun1();<\/code>1<\/sup>. If you are using a compiler that elides copies then all copies will be elided in both cases. If not eliding copies then the reference case will involve one less copy:<\/p>\n