Someone explain about reference_wrapper in c++

Question

I know a little about reference_wrapper in c++. I searched the internet and sololearn forum, but couldn't find a good explanation of it. Explain how it works, not just how to use it. Thank you ^^

Answer 1

If you look at the description of std::reference_wrapper,

https://en.cppreference.com/w/cpp/utility/functional/reference_wrapper

the main usage of it is " as a mechanism to store references inside standard containers (like std::vector) which cannot normally hold references...".

Sure enough, standard containers including std::vector cannot hold references. If you do something like

std::vector<int&> vec;

, the compiler complains. An explanation of why this is not allowed can be found here:

https://stackoverflow.com/questions/922360/why-cant-i-make-a-vector-of-references

For whatever reason if you need to store references in there, std::reference_wrapper can help construct such an object type compatible with standard containers. See:

https://code.sololearn.com/cm8tyNEq2usa/?ref=app

Answer 2

std::reference_wapper::get is documented here:

http://www.cplusplus.com/reference/functional/reference_wrapper/get/

Basically, it returns the referred element, which allows you to assign values to it. In most circumstances (including std::cout) you would want to use it, since trying to print the reference_wrapper object itself depends on whether the type of the referred element has a non-templated operator<<.

https://stackoverflow.com/questions/34144326/why-reference-wrapper-behaves-differently-for-built-in-types

This is an example where std::cout fails for std::reference_wrapper<std::string>, but works when using the .get method which returns the actual string variable.

Answer 3

Thank you! I got it, wish you a happy day ahead ^-^

Answer 4

Hmm... I think I get it, but I have one doubt. When I use this line (in the code you attached), "std::cout<<obj[0];", it prints 39(correct output)
But when I use "std::cout<<obj[0].get();', it still prints 39(correct output). So what's happening here? For assigning a new value to a reference, I have to use get() method, else it won't work. But how get() method does nothing in std::cout'ing it?