When this temporary object is destoryed?Before entry the function or after the function returned?

Question

When this temporary object(CTest()) is destoryed?Before entry the function(fooFunc) or after the function(fooFunc) returned?

I know the code below is not right indeed(Thank you,asmmo. ). The reason is that:

"The idea is that a function taking a non-const reference parameter is stating that it wants to modify the parameter and allowing it to go back to the caller. Doing so with a temporary is meaningless and most likely an error."

I would be thankful for any hint about this question.

class CTest;
void fooFunc(CTest&){};
fooFunc(CTest());

What do you mean with a temporary object. Do you have a piece of example code to demonstrate what you try to achieve? — Rennnyyy, May 23 '20 at 10:59
it can be passed as an rvalue reference or as an lvalue reference to const-qualified type — user7860670, May 23 '20 at 11:02
https://stackoverflow.com/questions/1565600/how-come-a-non-const-reference-cannot-bind-to-a-temporary-object?noredirect=1&lq=1 — asmmo, May 23 '20 at 11:02
Does this answer your question? [Can't pass temporary object as reference](https://stackoverflow.com/questions/27463785/cant-pass-temporary-object-as-reference) — asmmo, May 23 '20 at 11:03
@asmmo Thank you for your help.The url is related.But i have some doubt,so i send the feedback(no) to you. — sunshilong369, May 23 '20 at 11:29
A standards compliant compiler shouldn't compile your code: https://godbolt.org/z/oeUuwm. Visual studio does but its wrong — Alan Birtles, May 23 '20 at 11:34
@Alan Birtles You are right indeed.See my question again.I have posted the reason.I want to know that when the temporary object is destroyed if it's right. — sunshilong369, May 23 '20 at 11:35
The answer is the standard doesn't allow this behaviour so doesn't specify how it works — Alan Birtles, May 23 '20 at 11:39

score 2 · Accepted Answer · answered May 23 '20 at 11:40

2

Temporaries are destroyed at the end of full-expression. In your case this means that CTest() will be destroyed after fooFunc returns.

Note, that your example is ill-formed. fooFunc's parameter should be an rvalue reference or a const lvalue reference. Supposedly you use Visual Studio. Pass the /W4 flag, so VS will emit a warning that you use a nonstandard extension (so you know about that you use something which won't compile for other compilers).

answered May 23 '20 at 11:40

geza

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Is it stated in the C++ standard that temporaries are destroyed at the end of full-expression? – sunshilong369 May 23 '20 at 11:42
@sunshilong369: yes, here are the exact rules: http://eel.is/c++draft/class.temporary#4, http://eel.is/c++draft/class.temporary#5 and http://eel.is/c++draft/class.temporary#6 – geza May 23 '20 at 11:45
At the end of full-expression？ Do you mean after `function(fooFunc)` returned? – sunshilong369 May 23 '20 at 13:03
@sunshilong369: yes, this is the second sentence of my answer :) – geza May 23 '20 at 13:23

When this temporary object is destoryed?Before entry the function or after the function returned?

1 Answers1