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I have a function:
void AddImage(const Image &im);This function does not need image to be modifiable, but it stores the image a const reference to be used later. Thus, this function should not allow temporaries. Without any precaution the following works:
Image GetImage(); ... AddImage(GetImage());Is there a way to prevent this function call?
2
There are couple of mechanisms that prevent passing temporary to a const ref parameter.
Delete the overload with r-value parameter: void AddImage(const Image &&) = delete;
Use const pointer: void AddImage(const Image*) . This method will work pre-C++11
Use reference wrapper: void AddImage(std::reference_wrapper<const Image>)
First method should be preferred when using C++11 supporting compiler. It makes the intention clear. Second method requires a runtime check of nullptr and does not convey the idea that the image cannot be null. Third method works and makes the intention clear, however, it is too explicit.
6 Comments
Now I can detect problems with passing temporaries at compile-time. Thanks a lot!
2019-03-23T19:34:33.8Z+00:00
I would really appreciate if you could elaborate more on how std::reference_wrapper works in this case. What happens if the arguments is a temporary obj std::string, and the function returns a pointer to its .data()? Will it work in this case?
2020-04-21T08:30:57.963Z+00:00
I think (not sure) 3rd option uses a constructor with r-value overload deleted. Thus it will prevent temporary to be passed as a reference.
2020-04-22T09:17:06.743Z+00:00
Note that option 1 requires linearly (or exponential, if you want to avoid ambiguity) many overloads if you have multiple const reference arguments which you need prevent temporaries from being passed.
2020-06-17T18:26:22.793Z+00:00
What is meant by the third method being "too explicit"?
2021-02-25T19:05:32.64Z+00:00
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